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Full text of "Physics Volume 1 (Std11 - English Medium)"

PHYSICS 

HIGHER SECONDARY 
FIRST YEAR 

VOLUME - I 



Revised based on the recommendation of the 
Textbook Development Committee 



Untouchability is a sin 
Untouchability is a crime 
Untouchability is inhuman 




< **w£* 



TAMILNADU TEXTBOOK 
CORPORATION 

COLLEGE ROAD, CHENNAI - 600 006 



© Government of Tamilnadu 
First edition - 2004 
Revised edition - 2007 



CHAIRPERSON 

Dr. S. GUNASEKARAN 

Reader 

Post Graduate and Research Department of Physics 

Pachaiyappa's College, Chennai - 600 030 



Re v i ewers 

P. 5ARVAJ ANA RAJ AN 
Selection Grade Lecturer in Physics 
Govt. Art s Col I ege 
Nandanam, Chennai - 600 035 

5. KEMASARI 

Selection Grade Lecturer in Physics 
Queen Mary's College ( Autonomous) 
Chennai - 600 004 

Dr. K. MANIMEGALAI 
Reader ( Physi cs) 
The Ethi raj Col I ege for Wo men 
Chennai - 600 008 



Aut hors 

S. PONNUSAMY 

Asst. Professor of Physi cs 

S. R. M. Engi neeri ng Col I ege 

S. R. M I nsti tute of Sci ence and Technol ogy 

( Deerred Uni versi ty) 

Katt ankul at hur - 603 203 



S. RASARASAN 

P. G. Assi st ant i n Physi cs 

Govt. Hr. Sec. School 

Kodambakkam, Chennai - 600 024 

Gl Rl J A RAMANUJ AM 
P. G. Assi st ant i n Physi cs 
Govt. Girls' Hr . Sec. School 
Ashok Nagar , Chennai - 600 083 

P. LOGANATHAN 
P. G. Assi st ant i n Physi cs 
Govt. Oris' Hr . Sec. School 
Ti ruchengode - 637 211 
Namakkal District 

Dr.R. RAJ KUMAR 

P. G. Assi st ant i n Physics 

Dharmamurthi Rao Bahadur Ca I aval a 

Cunnan Chetty' s Hr . Sec. School 
Chennai - 600 Oil 



Dr.N. VIJAYA N 
P r i nci p d 

Zi on Matri c Hr . Sec. 

Sel ai yur 

Chennai - 600 073 



School 



Price Rs. 



This book has been prepared by the Directorate of School Education 
on behalf of the Government of Tarrilnadu 



The book has been printed on 60 GSM paper 



Preface 

The most important and crucial stage of school education is the 
higher secondary level. This is the transition level from a generalised 
curriculum to a discipline-based curriculum. 

In order to pursue their career in basic sciences and professional 
courses, students take up Physics as one of the subjects. To provide 
them sufficient background to meet the challenges of academic and 
professional streams, the Physics textbook for Std. XI has been reformed, 
updated and designed to include basic information on all topics. 

Each chapter starts with an introduction, followed by subject matter. 
All the topics are presented with clear and concise treatments. The 
chapters end with solved problems and self evaluation questions. 

Understanding the concepts is more important than memorising. 
Hence it is intended to make the students understand the subject 
thoroughly so that they can put forth their ideas clearly. In order to 
make the learning of Physics more interesting, application of concepts 
in real life situations are presented in this book. 

Due importance has been given to develop in the students, 
experimental and observation skills. Their learning experience would 
make them to appreciate the role of Physics towards the improvement 
of our society. 

The following are the salient features of the text book. 

• The data has been systematically updated. 

• Figures are neatly presented. 

• Self- evaluation questions (only samples) are included to sharpen 
the reasoning ability of the student. 

• As Physics cannot be understood without the basic knowledge 
of Mathematics, few basic ideas and formulae in Mathematics 
are given. 

While preparing for the examination, students should not 
restrict themselves, only to the questions/problems given in the 
self evaluation. They must be prepared to answer the questions 
and problems from the text/syllabus. 

Sincere thanks to Indian Space Research Organisation (ISRO) for 
providing valuable information regarding the Indian satellite programme. 

- Dr. S. Gunasekaran 

Chairperson 



SYLLABUS (180 periods) 

UNIT - 1 Nature of the Physical World and Measurement (7 periods) 

Physics - scope and excitement - physics in relation to technology 
and society. 

Forces in nature - gravitational, electromagnetic and nuclear 
forces (qualitative ideas) 

Measurement - fundamental and derived units - length, mass 
and time measurements. 

Accuracy and precision of measuring instruments, errors in 
measurement - significant figures. 

Dimensions - dimensions of physical quantities - dimensional 
analysis - applications. 

UNIT - 2 Kinematics (29 periods) 

Motion in a straight line - position time graph - speed and 
velocity - uniform and non-uniform motion - uniformly accelerated 
motion - relations for uniformly accelerated motions. 

Scalar and vector quantities - addition and subtraction of vectors, 
unit vector, resolution of vectors - rectangular components, 
multiplication of vectors - scalar, vector products. 

Motion in two dimensions - projectile motion - types of projectile 

- horizontal and oblique projectile. 

Force and inertia, Newton's first law of motion. 

Momentum - Newton's second law of motion - unit of force - 
impulse. 

Newton's third law of motion - law of conservation of linear 
momentum and its applications. 

Equilibrium of concurrent forces - triangle law, parallelogram 
law and Lami's theorem - experimental proof. 

Uniform circular motion - angular velocity - angular acceleration 

- relation between linear and angular velocities. Centripetal force - 
motion in a vertical circle - bending of cyclist - vehicle on level circular 
road - vehicle on banked road. 

Work done by a constant force and a variable force - unit of 
work. 



Energy - Kinetic energy, work - energy theorem - potential energy 

- power. 

Collisions - Elastic and in-elastic collisions in one dimension. 

UNIT - 3 Dynamics of Rotational Motion (14 periods) 

Centre of a two particle system - generalization - applications - 
equilibrium of bodies, rigid body rotation and equations of rotational 
motion. Comparison of linear and rotational motions. 

Moment of inertia and its physical significance - radius of gyration 

- Theorems with proof, Moment of inertia of a thin straight rod, circular 
ring, disc cylinder and sphere. 

Moment of force, angular momentum. Torque - conservation of 
angular momentum. 

UNIT - 4 Gravitation and Space Science (16 periods) 

The universal law of gravitation; acceleration due to gravity and 
its variation with the altitude, latitude, depth and rotation of the Earth. 

- mass of the Earth. Inertial and gravitational mass. 

Gravitational field strength - gravitational potential - gravitational 
potential energy near the surface of the Earth - escape velocity - 
orbital velocity - weightlessness - motion of satellite - rocket propulsion 

- launching a satellite - orbits and energy. Geo stationary and polar 
satellites - applications - fuels used in rockets - Indian satellite 
programme. 

Solar system - Helio, Geo centric theory - Kepler's laws of planetary 
motion. Sun - nine planets - asteroids - comets - meteors - meteroites 

- size of the planets - mass of the planet - temperature and atmosphere. 

Universe - stars - constellations - galaxies - Milky Way galaxy - 
origin of universe. 

UNIT - 5 Mechanics of Solids and Fluids (18 periods) 

States of matter- inter-atomic and inter-molecular forces. 

Solids - elastic behaviour, stress - strain relationship, Hooke's 
law - experimental verification of Hooke's law - three types of moduli 
of elasticity - applications (crane, bridge). 

Pressure due to a fluid column - Pascal's law and its applications 
(hydraulic lift and hydraulic brakes) - effect of gravity on fluid pressure. 



Surface energy and surface tension, angle of contact - application 
of surface tension in (i) formation of drops and bubbles (ii) capillary 
rise (iii) action of detergents. 

Viscosity - Stoke's law - terminal velocity, streamline flow - 
turbulant flow - Reynold's number - Bernoulli's theorem - applications 

- lift on an aeroplane wing. 

UNIT - 6 Oscillations (12 periods) 

Periodic motion - period, frequency, displacement as a function 
of time. 

Simple harmonic motion - amplitude, frequency, phase - uniform 
circular motion as SHM. 

Oscillations of a spring, liquid column and simple pendulum - 
derivation of expression for time period - restoring force - force constant. 
Energy in SHM. kinetic and potential energies - law of conservation of 
energy. 

Free, forced and damped oscillations. Resonance. 

UNIT - 7 Wave Motion (17 periods) 

Wave motion- longitudinal and transverse waves - relation 
between v, n, X. 

Speed of wave motion in different media - Newton's formula - 
Laplace's correction. 

Progressive wave - displacement equation -characteristics. 

Superposition principle, Interference - intensity and sound level 

- beats, standing waves (mathematical treatment) - standing waves in 
strings and pipes - sonometer - resonance air column - fundamental 
mode and harmonics. 

Doppler effect (quantitative idea) - applications. 

UNIT - 8 Heat and Thermodynamics (17 periods) 

Kinetic theory of gases - postulates - pressure of a gas - kinetic 
energy and temperature - degrees of freedom (mono atomic, diatomic 
and triatomic) - law of equipartition of energy - Avogadro's number. 

Thermal equilibrium and temperature (zeroth law of 
thermodynamics) Heat, work and internal energy. Specific heat - specific 



heat capacity of gases at constant volume and pressure. Relation 
between C and C . 

P v 

First law of thermodynamics - work done by thermodynamical 
system - Reversible and irreversible processes - isothermal and adiabatic 
processes - Carnot engine - refrigerator - efficiency - second law of 
thermodynamics. 

Transfer of heat - conduction, convection and radiation - Thermal 
conductivity of solids - black body radiation - Prevost's theory - Kirchoff s 
law - Wien's displacement law, Stefan's law (statements only). Newton's 
law of cooling - solar constant and surface temperature of the Sun- 
pyrheliometer . 

UNIT - 9 Ray Optics (16 periods) 

Reflection of light - reflection at plane and curved surfaces. 

Total internal refelction and its applications - determination of 
velocity of light - Michelson's method. 

Refraction - spherical lenses - thin lens formula, lens makers 
formula - magnification - power of a lens - combination of thin lenses 
in contact. 

Refraction of light through a prism - dispersion - spectrometer - 
determination of \i - rainbow. 

UNIT - 10 Magnetism (10 periods) 

Earth's magnetic field and magnetic elements. Bar magnet - 
magnetic field lines 

Magnetic field due to magnetic dipole (bar magnet) along the axis 
and perpendicular to the axis. 

Torque on a magnetic dipole (bar magnet) in a uniform magnetic 
field. 

Tangent law - Deflection magnetometer - Tan A and Tan B 
positions. 

Magnetic properties of materials - Intensity of magnetisation, 
magnetic susceptibility, magnetic induction and permeability 

Dia, Para and Ferromagnetic substances with examples. 

Hysteresis. 



EXPERIMENTS (12 x 2 = 24 periods) 

1. To find the density of the material of a given wire with the help 
of a screw gauge and a physical balance. 

2. Simple pendulum - To draw graphs between (i) L and T and 
(ii) L and T 2 and to decide which is better. Hence to determine the 
acceleration due to gravity. 

3. Measure the mass and dimensions of (i) cylinder and (ii) solid 
sphere using the vernier calipers and physical balance. Calculate 
the moment of inertia. 

4. To determine Young's modulus of the material of a given wire by 
using Searles' apparatus. 

5. To find the spring constant of a spring by method of oscillations. 

6. To determine the coefficient of viscosity by Poiseuille's flow method. 

7. To determine the coefficient of viscosity of a given viscous liquid 
by measuring the terminal velocity of a given spherical body. 

8. To determine the surface tension of water by capillary rise method. 

9. To verify the laws of a stretched string using a sonometer. 

10. To find the velocity of sound in air at room temperature using the 
resonance column apparatus. 

11. To determine the focal length of a concave mirror 

12. To map the magnetic field due to a bar magnet placed in the 
magnetic meridian with its (i) north pole pointing South and 
(ii) north pole pointing North and locate the null points. 



( 


CONTENTS 


Page No. 




Mathematical Notes 


1 


1. 


Nature of the Physical World 






and Measurement 


13 


2. 


Kinematics 


37 


3. 


Dynamics of Rotational Motion 


120 


4. 


Gravitation and Space Science 


149 


5. 


Mechanics of Solids and Fluids 


194 




Annexure 


237 




Logarithmic and other tables 


252 


V 


(Unit 6 to 10 continues in Volume II) 





1. Nature of the 
Physical World and Measurement 

The history of humans reveals that they have been making 
continuous and serious attempts to understand the world around them. 
The repetition of day and night, cycle of seasons, volcanoes, rainbows, 
eclipses and the starry night sky have always been a source of wonder 
and subject of thought. The inquiring mind of humans always tried to 
understand the natural phenomena by observing the environment 
carefully. This pursuit of understanding nature led us to today's modern 
science and technology. 

1 . 1 Physics 

The word science comes from a Latin word "scientia" which means 
'to know'. Science is nothing but the knowledge gained through the 
systematic observations and experiments. Scientific methods include 
the systematic observations, reasoning, modelling and theoretical 
prediction. Science has many disciplines, physics being one of them. 
The word physics has its origin in a Greek word meaning 'nature'. 
Physics is the most basic science, which deals with the study of nature 
and natural phenomena. Understanding science begins with 
understanding physics. With every passing day, physics has brought to 
us deeper levels of understanding of nature. 

Physics is an empirical study. Everything we know about physical 
world and about the principles that govern its behaviour has been 
learned through observations of the phenomena of nature. The ultimate 
test of any physical theory is its agreement with observations and 
measurements of physical phenomena. Thus physics is inherently a 
science of measurement. 

1.1.1 Scope of Physics 

The scope of physics can be understood if one looks at its 
various sub-disciplines such as mechanics, optics, heat and 
thermodynamics, electrodynamics, atomic physics, nuclear physics, etc. 

13 



Mechanics deals with motion of particles and general systems of particles. 
The working of telescopes, colours of thin films are the topics dealt in 
optics. Heat and thermodynamics deals with the pressure - volume 
changes that take place in a gas when its temperature changes, working 
of refrigerator, etc. The phenomena of charged particles and magnetic 
bodies are dealt in electrodynamics. The magnetic field around a current 
carrying conductor, propagation of radio waves etc. are the areas where 
electrodynamics provide an answer. Atomic and nuclear physics deals 
with the constitution and structure of matter, interaction of atoms and 
nuclei with electrons, photons and other elementary particles. 

Foundation of physics enables us to appreciate and enjoy things 
and happenings around us. The laws of physics help us to understand 
and comprehend the cause-effect relationships in what we observe. 
This makes a complex problem to appear pretty simple. 

Physics is exciting in many ways. To some, the excitement comes 
from the fact that certain basic concepts and laws can explain a range 
of phenomena. For some others, the thrill lies in carrying out new 
experiments to unravel the secrets of nature. Applied physics is even 
more interesting. Transforming laws and theories into useful applications 
require great ingenuity and persistent effort. 

1.1.2 Physics, Technology and Society 

Technology is the application of the doctrines in physics for 
practical purposes. The invention of steam engine had a great impact 
on human civilization. Till 1933, Rutherford did not believe that energy 
could be tapped from atoms. But in 1938, Hann and Meitner discovered 
neutron-induced fission reaction of uranium. This is the basis of nuclear 
weapons and nuclear reactors. The contribution of physics in the 
development of alternative resources of energy is significant. We are 
consuming the fossil fuels at such a very fast rate that there is an 
urgent need to discover new sources of energy which are cheap. 
Production of electricity from solar energy and geothermal energy is a 
reality now, but we have a long way to go. Another example of physics 
giving rise to technology is the integrated chip, popularly called as IC. 
The development of newer ICs and faster processors made the computer 
industry to grow leaps and bounds in the last two decades. Computers 
have become affordable now due to improved production techniques 

14 



and low production costs. 

The legitimate purpose of technology is to serve poeple. Our society 
is becoming more and more science-oriented. We can become better 
members of society if we develop an understanding of the basic laws of 
physics. 

1.2 Forces of nature 

Sir Issac Newton was the first one to give an exact definition for 
force. 

"Force is the external agency applied on a body to change its state 
of rest and motion". 

There are four basic forces in nature. They are gravitational force, 
electromagnetic force, strong nuclear force and weak nuclear force. 

Gravitational force 

It is the force between any two objects in the universe. It is an 
attractive force by virtue of their masses. By Newton's law of gravitation, 
the gravitational force is directly proportional to the product of the 
masses and inversely proportional to the square of the distance between 
them. Gravitational force is the weakest force among the fundamental 
forces of nature but has the greatest large-scale impact on the universe. 
Unlike the other forces, gravity works universally on all matter and 
energy, and is universally attractive. 

Electromagnetic force 

It is the force between charged particles such as the force between 
two electrons, or the force between two current carrying wires. It is 
attractive for unlike charges and repulsive for like charges. The 
electromagnetic force obeys inverse square law. It is very strong compared 
to the gravitational force. It is the combination of electrostatic and 
magnetic forces. 

Strong nuclear force 

It is the strongest of all the basic forces of nature. It, however, 
has the shortest range, of the order of 10 15 m. This force holds the 
protons and neutrons together in the nucleus of an atom. 



15 



Weak nuclear force 

Weak nuclear force is important in certain types of nuclear process 
such as p-decay. This force is not as weak as the gravitational force. 

1.3 Measurement 

Physics can also be defined as the branch of science dealing with 
the study of properties of materials. To understand the properties of 
materials, measurement of physical quantities such as length, mass, 
time etc., are involved. The uniqueness of physics lies in the measurement 
of these physical quantities. 

1.3.1 Fundamental quantities and derived quantities 

Physical quantities can be classified into two namely, fundamental 
quantities and derived quantities. Fundamental quantities are quantities 
which cannot be expressed in terms of any other physical quantity. For 
example, quantities like length, mass, time, temperature are fundamental 
quantities. Quantities that can be expressed in terms of fundamental 
quantities are called derived quantities. Area, volume, density etc. are 
examples for derived quantities. 

1.3.2 Unit 

To measure a quantity, we always compare it with some reference 
standard. To say that a rope is 10 metres long is to say that it is 10 
times as long as an object whose length is defined as 1 metre. Such a 
standard is called a unit of the quantity. 

Therefore, unit of a physical quantity is defined as the established 
standard used for comparison of the given physical quantity. 

The units in which the fundamental quantities are measured are 
called fundamental units and the units used to measure derived quantities 
are called derived units. 

1 .3.3 System. International de Units (SI system of units) 

In earlier days, many system of units were followed to measure 
physical quantities. The British system of foot-pound-second or fps 
system, the Gaussian system of centimetre - gram - second or cgs 
system, the metre-kilogram - second or the mks system were the three 

16 



systems commonly followed. To bring uniformity, the General Conference 
on Weights and Measures in the year 1960, accepted the SI system of 
units. This system is essentially a modification over mks system and is, 
therefore rationalised mksA (metre kilogram second ampere) system. 
This rationalisation was essential to obtain the units of all the physical 
quantities in physics. 

In the SI system of units there are seven fundamental quantities 
and two supplementary quantities. They are presented in Table 1.1. 

Table 1.1 SI system of units 



Physical quantity 


Unit 


Symbol 


Fundamental quantities 






Length 


metre 


m 


Mass 


kilogram 


kg 


Time 


second 


s 


Electric current 


ampere 


A 


Temperature 


kelvin 


K 


Luminous intensity 


candela 


cd 


Amount of substance 


mole 


mol 


Supplementary quantities 






Plane angle 


radian 


rad 


Solid angle 


steradian 


sr 



1 .3.4 Uniqueness of SI system 

The SI system is logically far superior to all other systems. The 
SI units have certain special features which make them more convenient 
in practice. Permanence and reproduceability are the two important 
characteristics of any unit standard. The SI standards do not vary with 
time as they are based on the properties of atoms. Further SI system 
of units are coherent system of units, in which the units of derived 
quantities are obtained as multiples or submultiples of certain basic units. 
Table 1.2 lists some of the derived quantities and their units. 



17 



Table 1.2 Derived quantities and their units 



Physical Quantity 


Expression 


Unit 


Area 


length x breadth 


2 

m z 


Volume 


area x height 


m 3 


Velocity 


displacement/ time 


m s" 1 


Acceleration 


velocity / time 


_2 

m s z 


Angular velocity 


angular displacement / time 


rad s" 1 


Angular acceleration 


angular velocity / time 


rad s~ 2 


Density 


mass / volume 


kg nr 3 


Momentum 


mass x velocity 


kg m s _1 


Moment of intertia 


mass x (distance) 2 


kg m 2 


Force 


mass x acceleration 


kg m s -2 or N 


Pressure 


force / area 


N m" 2 or Pa 


Energy (work) 


force x distance 


N m or J 


Impulse 


force x time 


N s 


Surface tension 


force / length 


N m" 1 


Moment of force (torque) 


force x distance 


N m 


Electric charge 


current x time 


A s 


Current density 


current / area 


A m" 2 


Magnetic induction 


force / (current x length) 


N A" 1 m" 1 



1.3.5 SI standards 
Length 

Length is defined as the distance between two points. The SI unit 
of length is metre. 

One standard metre is equal to 1 650 763. 73 wavelengths of the 
orange - red light emitted by the individual atoms of krypton - 86 in a 
krypton discharge lamp. 

Mass 

Mass is the quantity of matter contained in a body. It is 
independent of temperature and pressure. It does not vary from place 

18 



to place. The SI unit of mass is kilogram. 

The kilogram is equal to the mass of the international prototype of 
the kilogram (a plantinum - iridium alloy cylinder) kept at the International 
Bureau of Weights and Measures at Sevres, near Paris, France. 

An atomic standard of mass has not yet been adopted because it 
is not yet possible to measure masses on an atomic scale with as much 
precision as on a macroscopic scale. 

Time 

Until 1960 the standard of time was based on the mean solar day, 
the time interval between successive passages of the sun at its highest 
point across the meridian. It is averaged over an year. In 1967, an 
atomic standard was adopted for second, the SI unit of time. 

One standard second is defined as the time taken for 
9 192 631 770 periods of the radiation corresponding to unperturbed 
transition between hyperfine levels of the ground state of cesium -133 
atom. Atomic clocks are based on this. In atomic clocks, an error of one 
second occurs only in 5000 years. 

Ampere 

The ampere is the constant current which, flowing through two straight 
parallel infinitely long conductors of negligible cross-section, and placed in 
vacuum 1 m apart, would produce between the conductors a force of 
2 x 10 ~ 7 newton per unit length of the conductors. 

Kelvin 

1 
The Kelvin is the fraction of of the thermodynamic 

temperature of the triple point of water*. 

Candela 

The candela is the luminous intensity in a given direction due to a 

* Triple point of water is the temperature at which saturated water vapour, 
pure water and melting ice are all in equilibrium. The triple point temperature of 
water is 273.16 K. 



19 



source, which emits monochromatic radiation of frequency 540 x 10 12 Hz 
and of which the radiant intensity in that direction is —r watt per steradian. 

Mole 

The mole is the amount of substance which contains as many 
elementary entities as there are atoms in 0.012 kg of carbon- 12. 

1 .3.6 Rules and conventions for writing SI units and their symbols 

1. The units named after scientists are not written with a capital 
initial letter. 

For example : newton, henry, watt 

2. The symbols of the units named after scientist should be written 
by a capital letter. 

For example : N for newton, H for henry, W for watt 

3. Small letters are used as symbols for units not derived from a 
proper name. 

For example : m for metre, kg for kilogram 

4. No full stop or other punctuation marks should be used within 
or at the end of symbols. 

For example : 50 m and not as 50 m. 

5. The symbols of the units do not take plural form. 
For example : 10 kg not as 10 kgs 

6. When temperature is expressed in kelvin, the degree sign is 
omitted. 

For example : 273 K not as 273° K 

(If expressed in Celsius scale, degree sign is to be included. For 
example 100° C and not 100 C) 

7. Use of solidus is recommended only for indicating a division of 
one letter unit symbol by another unit symbol. Not more than one 
solidus is used. 

For example : m s _1 or m / s, J / Kmol or J K _1 mol -1 but not 
J / K / mol. 



20 



8. Some space is always to be left between the number and the 
symbol of the unit and also between the symbols for compound units 
such as force, momentum, etc. 

For example, it is not correct to write 2.3m. The correct 
representation is 2.3 m; kg m s" 2 and not as kgms" 2 . 

9. Only accepted symbols should be used. 

For example : ampere is represented as A and not as amp. or am ; 
second is represented as s and not as sec. 

10. Numerical value of any physical quantity should be expressed 
in scientific notation. 

For an example, density of mercury is 1.36 x 10 4 kgrn 3 and not 
as 13600 kg m 3 . 

1.4 Expressing larger and smaller physical quantities 

Once the fundamental Table 1.3 Prefixes for power of ten 

units are defined, it is easier 
to express larger and smaller 
units of the same physical 
quantity. In the metric (SI) 
system these are related to the 
fundamental unit in multiples 
of 10 or 1/10. Thus 1 km is 
1000 m and 1 mm is 1/1000 
metre. Table 1.3 lists the 
standard SI prefixes, their 
meanings and abbreviations. 

In order to measure very 
large distances, the following 
units are used. 

(i) Light year 

Light year is the distance 
travelled by light in one year 
in vacuum. 



Power of ten 


Prefix 


Abbreviation 


10-1 5 


femto 


f 


10-1 2 


pico 


P 


10 9 


nano 


n 


10 6 


micro 


H 


10 3 


milli 


m 


lO- 2 


centi 


c 


io-i 


deci 


d 


10 1 


deca 


da 


10 2 


hecto 


h 


10 3 


kilo 


k 


10 6 


mega 


M 


10 9 


giga 


G 


10 12 


tera 


T 


10 15 


peta 


P 



21 



Distance travelled = velocity of light x 1 year 

.". 1 light year = 3xl0 8 ms~ 1 xl year (in seconds) 

= 3 x 10 8 x 365.25 x 24 x 60 x 60 

= 9.467 x 10 15 m 
1 light year = 9.467 x 10 15 m 

(ii) Astronomical unit 

Astronomical unit is the mean distance of the centre of the Sun 
from the centre of the Earth. 

1 Astronomical unit (AU) = 1.496 x 10 11 m 

1.5 Determination of distance 

For measuring large distances such as the distance of moon or 
a planet from the Earth, special methods are adopted. Radio-echo 
method, laser pulse method and parallax method are used to determine 
very large distances. 

Laser pulse method 

The distance of moon from the Earth can be determined using 
laser pulses. The laser pulses are beamed towards the moon from a 
powerful transmitter. These pulses are reflected back from the surface 
of the moon. The time interval between sending and receiving of the 
signal is determined very accurately. 

If t is the time interval and c the velocity of the laser pulses, then 

Ct 

the distance of the moon from the Earth is d = — . 

1.6 Determination of mass 

The conventional method of finding the mass of a body in the 
laboratory is by physical balance. The mass can be determined to an 
accuracy of 1 mg. Now-a-days, digital balances are used to find the 
mass very accurately. The advantage of digital balance is that the mass 
of the object is determined at once. 

1.7 Measurement of time 

We need a clock to measure any time interval. Atomic clocks provide 
better standard for time. Some techniques to measure time interval are 
given below. 

22 



Quartz clocks 

The piezo-electric property* of a crystal is the principle of quartz 
clock. These clocks have an accuracy of one second in every 10 9 seconds. 

Atomic clocks 

These clocks make use of periodic vibration taking place within 
the atom. Atomic clocks have an accuracy of 1 part in 10 13 seconds. 

1.8 Accuracy and precision of measuring instruments 

All measurements are made with the help of instruments. The 
accuracy to which a measurement is made depends on several factors. 
For example, if length is measured using a metre scale which has 
graduations at 1 mm interval then all readings are good only upto this 
value. The error in the use of any instrument is normally taken to be half 
of the smallest division on the scale of the instrument. Such an error is 
called instrumental error. In the case of a metre scale, this error is 
about 0.5 mm. 

Physical quantities obtained from experimental observation always 
have some uncertainity. Measurements can never be made with absolute 
precision. Precision of a number is often indicated by following it with 
+ symbol and a second number indicating the maximum error likely. 

For example, if the length of a steel rod = 56.47 +3 mm then the 
true length is unlikely to be less than 56.44 mm or greater than 
56.50 mm. If the error in the measured value is expressed infraction, it 
is called fractional error and if expressed in percentage it is called 
percentage error. For example, a resistor labelled "470 Q, 10%" probably 
has a true resistance differing not more than 10% from 470 Q. So the 
true value lies between 423 D. and 517 Q. 

1.8.1 Significant figures 

The digits which tell us the number of units we are reasonably 
sure of having counted in making a measurement are called significant 
figures. Or in other words, the number of meaningful digits in a number 
is called the number of significant figures. A choice of change of different 
units does not change the number of significant digits or figures in a 
measurement. 

* When pressure is applied along a particular axis of a crystal, an electric 
potential difference is developed in a perpendicular axis. 

23 



For example, 2.868 cm has four significant figures. But in different 
units, the same can be written as 0.02868 m or 28.68 mm or 28680 
(am. All these numbers have the same four significant figures. 

From the above example, we have the following rules. 

i) All the non-zero digits in a number are significant. 

ii) All the zeroes between two non-zeroes digits are significant, 
irrespective of the decimal point. 

iii) If the number is less than 1 , the zeroes on the right of decimal 
point but to the left of the first non-zero digit are not significant. (In 
0.02868 the underlined zeroes are not significant). 

iv) The zeroes at the end without a decimal point are not 
significant. (In 23080 (am, the trailing zero is not significant). 

v) The trailing zeroes in a number with a decimal point are 
significant. (The number 0.07100 has four significant digits). 

Examples 

i) 30700 has three significant figures, 
ii) 132.73 has five significant figures, 
iii) 0.00345 has three and 
iv) 40.00 has four significant figures. 

1.8.2 Rounding off 

Calculators are widely used now-a-days to do the calculations. 
The result given by a calculator has too many figures. In no case the 
result should have more significant figures than the figures involved in 
the data used for calculation. The result of calculation with number 
containing more than one uncertain digit, should be rounded off. The 
technique of rounding off is followed in applied areas of science. 

A number 1.876 rounded off to three significant digits is 1.88 
while the number 1.872 would be 1.87. The rule is that if the insignificant 
digit (underlined) is more than 5, the preceeding digit is raised by 1, 
and is left unchanged if the former is less than 5. 

If the number is 2.845, the insignificant digit is 5. In this case, 
the convention is that if the preceeding digit is even, the insignificant 
digit is simply dropped and, if it is odd, the preceeding digit is raised 
by 1. Following this, 2.845 is rounded off to 2.84 where as 2.815 is 
rounded off to 2.82. 

24 



Examples 

1. Add 17.35 kg, 25.8 kg and 9.423 kg. 

Of the three measurements given, 25.8 kg is the least accurately 
known. 

.-. 17.35 + 25.8 + 9.423 = 52.573 kg 
Correct to three significant figures, 52.573 kg is written as 
52.6 kg 

2. Multiply 3.8 and 0.125 with due regard to significant figures. 

3.8 x 0.125 = 0.475 

The least number of significant figure in the given quantities is 2. 
Therefore the result should have only two significant figures. 

.-. 3.8 x 0.125 = 0.475 = 0.48 

1.8.3 Errors in Measurement 

The uncertainity in the measurement of a physical quantity is 
called error. It is the difference between the true value and the measured 
value of the physical quantity. Errors may be classified into many 
categories. 

(i) Constant errors 

It is the same error repeated every time in a series of observations. 
Constant error is due to faulty calibration of the scale in the measuring 
instrument. In order to minimise constant error, measurements are 
made by different possible methods and the mean value so obtained is 
regarded as the true value. 

(ii) Systematic errors 

These are errors which occur due to a certain pattern or system. 
These errors can be minimised by identifying the source of error. 
Instrumental errors, personal errors due to individual traits and errors 
due to external sources are some of the systematic errors. 

(Hi) Gross errors 

Gross errors arise due to one or more than one of the following 
reasons. 

(1) Improper setting of the instrument. 

25 



(2) Wrong recordings of the observation. 

(3) Not taking into account sources of error and precautions. 

(4) Usage of wrong values in the calculation. 

Gross errros can be minimised only if the observer is very careful 
in his observations and sincere in his approach. 

(iv) Random errors 

It is very common that repeated measurements of a quantity give 
values which are slightly different from each other. These errors have 
no set pattern and occur in a random manner. Hence they are called 
random errors. They can be minimised by repeating the measurements 
many times and taking the arithmetic mean of all the values as the 
correct reading. 

The most common way of expressing an error is percentage error. 
If the accuracy in measuring a quantity x is Ax, then the percentage 

Ax 
error in x is given by — x 100 %. 

1.9 Dimensional Analysis 

Dimensions of a physical quantity are the powers to which the 
fundamental quantities must be raised. 

displacement 

We know that velocity = 

time 

[L] 
= [T] 

= [M^T- 1 ] 

where [M] , [L] and [T] are the dimensions of the fundamental quantities 
mass, length and time respectively. 

Therefore velocity has zero dimension in mass, one dimension in 
length and -1 dimension in time. Thus the dimensional formula for 
velocity is [M°L 1 T~ 1 ] or simply [LT _1 ].The dimensions of fundamental 
quantities are given in Table 1.4 and the dimensions of some derived 
quantities are given in Table 1.5 



26 



Table 1.4 Dimensions of fundamental quantities 



Fundamental quantity 


Dimension 


Length 


L 


Mass 


M 


Time 


T 


Temperature 


K 


Electric current 


A 


Luminous intensity 


cd 


Amount of subtance 


mol 



Table 1.5 Dimensional formulae of some derived quantities 



Physical quantity 


Expression 


Dimensional formula 


Area 


length x breadth 


[L 2 ] 


Density 


mass / volume 


[ML 3 ] 


Acceleration 


velocity / time 


[LT- 2 ] 


Momentum 


mass x velocity 


[MLT 1 ] 


Force 


mass x acceleration 


[MLT 2 ] 


Work 


force x distance 


[MI^T 2 ] 


Power 


work / time 


[MI^T 3 ] 


Energy 


work 


[MI^T 2 ] 


Impulse 


force x time 


[MLT 1 ] 


Radius of gyration 


distance 


[L] 


Pressure 


force / area 


[ML^T 2 ] 


Surface tension 


force / length 


[Mr 2 ] 


Frequency 


1 / time period 


[T 1 ] 


Tension 


force 


[MLT 2 ] 


Moment of force (or torque) 


force x distance 


[ML 2 ^ 2 ] 


Angular velocity 


angular displacement / time 


[T 1 ] 


Stress 


force / area 


[ML- l T 2 ] 


Heat 


energy 


[ML 2 ^ 2 ] 


Heat capacity 


heat energy/ temperature 


[ML^K 1 ] 


Charge 


current x time 


[AT] 


Faraday constant 


Avogadro constant x 






elementary charge 


[AT mol" 1 ] 


Magnetic induction 


force / (current x length) 


[MT 2 A' 1 ] 



27 



Dimensional quantities 

Constants which possess dimensions are called dimensional 
constants. Planck's constant, universal gravitational constant are 
dimensional constants. 

Dimensional variables are those physical quantities which possess 
dimensions but do not have a fixed value. Example - velocity, force, etc. 

Dimensionless quantities 

There are certain quantities which do not possess dimensions. 
They are called dimensionless quantities. Examples are strain, angle, 
specific gravity, etc. They are dimensionless as they are the ratio of two 
quantities having the same dimensional formula. 

Principle of homogeneity of dimensions 

An equation is dimensionally correct if the dimensions of the various 
terms on either side of the equation are the same. This is called the 
principle of homogeneity of dimensions. This principle is based on the 
fact that two quantities of the same dimension only can be added up, 
the resulting quantity also possessing the same dimension. 

The equation A + B = C is valid only if the dimensions of A, B and 
C are the same. 

1.9.1 Uses of dimensional analysis 

The method of dimensional analysis is used to 

(i) convert a physical quantity from one system of units to another. 

(ii) check the dimensional correctness of a given equation. 

(iii) establish a relationship between different physical quantities 
in an equation. 

(i) To convert a physical quantity from one system of units to another 

Given the value of G in cgs system is 6.67 x 10 8 dyne cm 2 g 2 . 



Calculate its value in SI 


units. 




In cgs system 




In SI system 


G cgs = 6.67 x 


io- 8 


G = ? 


M l = lg 




M 2 = 1 kg 


L 1 = 1 cm 




L 2 = lm 


T l = Is 




T 2 = Is 



28 



The dimensional formula for gravitational constant is \m^L 3 T' 

In cgs system, dimensional formula for G is I Mf l\ T : z 

In SI system, dimensional formula for G is I M| L| T| 
Here x = -1, y = 3, z = -2 

,. g[m 2 *L/T/] = G cgs [m^T*] 



or 



cgs 



6.67 x 10" 8 



M 2 




^ 2 


a 


T 2 



"is ~ 

1/cg 


-1 


"1 


zra 


3 


"1 s" 
_1 s_ 


-2 


_ 1 m _ 






is 


-1 


lcm I'M 

_100 cmj L J 




100( 


>9_ 



= 6.67 x 10" 

= 6.67 x 10 n 

In SI units, 

G = 6.67 x 10~ n N m 2 kg 2 

(ii) To check the dimensional correctness of a given equation 

Let us take the equation of motion 

s = ut + (V2)at 2 

Applying dimensions on both sides, 

[L] = [LT- 1 ] [T] + [UT- 2 ] [T 2 ] 

(Vfc is a constant having no dimension) 

[L] = [L] + [L] 

As the dimensions on both sides are the same, the equation is 
dimensionally correct. 

(Hi) To establish a relationship between the physical quantities 
in an equation 

Let us find an expression for the time period T of a simple pendulum. 
The time period T may depend upon (i) mass m of the bob (ii) length I 
of the pendulum and (iii) acceleration due to gravity g at the place where 
the pendulum is suspended. 



29 



(i.e) T a m x iy g z 

or T = k m x iy g z ...(1) 

where k is a dimensionless constant of propotionality. Rewriting 
equation (1) with dimensions, 

[T 1 ] = [M x ] [LV] [LT 2 ] Z 

IT 1 ] = ]M x Ly + zt 2z ] 

Comparing the powers of M, L and T on both sides 
x = 0, y+z=0 and -1z = 1 

Solving for x, y and z, x = 0, y = ¥2 and z = -V2 

From equation (1), T = k m° f 2 g- 1/2 



r m 



k i~ 9 



Experimentally the value of k is determined to be 2n. 

••■ T = ^fg 

1 .9.2 Limitations of Dimensional Analysis 

(i) The value of dimensionless constants cannot be determined 
by this method. 

(ii) This method cannot be applied to equations involving 
exponential and trigonometric functions. 

(iii) It cannot be applied to an equation involving more than three 
physical quantities. 

(iv) It can check only whether a physical relation is dimensionally 
correct or not. It cannot tell whether the relation is absolutely correct 

1 „ 
or not. For example applying this technique s = ut + — at z is dimensionally 

correct whereas the correct relation is s = ut + —at 2 . 



30 



Solved Problems 

1.1 A laser signal is beamed towards a distant planet from the Earth 
and its reflection is received after seven minutes. If the distance 
between the planet and the Earth is 6.3 x 10 10 m, calculate the 
velocity of the signal. 

Data : d = 6.3 x 10 10 m, t = 7 minutes = 7 x 60 = 420 s 

Solution : Ifd is the distance of the planet, then total distance travelled 
by the signal is 2d. 

2d 2x6.3xl0 10 _ n „ 8 _! 

.-. velocity = — = — = 3xlO b m s 

y t 420 

1.2 A goldsmith put a ruby in a box weighing 1.2 kg. Find the total 
mass of the box and ruby applying principle of significant figures. 
The mass of the ruby is 5.42 g. 

Data : Mass of box = 1.2 kg 

Mass of ruby = 5.42 g = 5.42 x 10~ 3 kg = 0.00542 kg 

Solution: Total mass = mass of box + mass of ruby 

= 1.2 + 0.00542 = 1.20542 kg 

After rounding off, total mass = 1.2 kg 

, h 

1.3 Check whether the equation ^ is dimensionally correct 

[X - wavelength, h - Planck's constant, m - mass, v - velocity). 
Solution: Dimension of Planck's constant h is [ML 2 T 1 ] 

Dimension ofX is [L] 

Dimension of mis [M] 

Dimension ofvis {LT~ 1 l 

Rewriting X = using dimension 

mv 



[ML 2 T- r 



[M^LT- 1 ] 

As the dimensions on both sides of the equation are same, the given 
equation is dimensionally correct. 

31 



1 .4 Multiply 2.2 and 0.225. Give the answer correct to significant figures. 

Solution : 2.2 x 0.225 = 0.495 

Since the least number of significant figure in the given data is 2, the 
result should also have only two significant figures. 

:. 2.2 x 0.225 = 0.50 

1.5 Convert 76 cm of mercury pressure into N m 2 using the method of 
dimensions. 

Solution : In cgs system, 76 cm of mercury 

pressure = 76 x 13.6 x 980 dyne cm~ 2 
Let this be P y Therefore P 1 = 76 x 13.6 x 980 dyne cm' 2 
In cgs system, the dimension of pressure is [M 1 a L 1 b T 1 c ] 
Dimension of pressure is [MLr 1 T~ 2 ]. Comparing this with [M 2 a L 2 b T 2 c ] 
we have a= 1, b = -1 and c = -2. 



Pressure in SI system 



P = P 



Mi 



^2 



ie P n = 76x13.6x980 



10" 3 kg 



10 2 m 
1 m 



- 1 -, -1-2 

Is 



lkg 

= 76 x 13.6 x 980 xl0~ 3 xlO 2 
= 101292.8 N m 2 
= 1.01 x 10 5 Nm 2 



Is 



32 



Self evaluation 

(The questions and problems given in this self evaluation are only samples. 
In the same way any question and problem could be framed from the text 
matter. Students must be prepared to answer any question and problem 
from the text matter, not only from the self evaluation.) 

1 . 1 Which of the following are equivalent? 

(a) 6400 km and 6.4 x 10 8 cm (b) 2 x 1 4 cm and 2 x 10 6 mm 
(c) 800 m and 80 x 10 2 m (d) 1 00 [im and 1 mm 

1 .2 Red light has a wavelength of 7000 A. In [im it is 
(a) 0. 7 y.m (b) 7 [im 
(c)70^im (d)0.07^im 

1.3 A speck of dust weighs 1.6 x 10~ 10 kg. How many such particles 
would weigh 1 . 6 kg? 

(a) 1CT 10 (b) 10 10 

(c) 10 (d) 10- 1 

1.4 The force acting on a particle is found to be proportional to velocity. 
The constant of proportionality is measured in terms of 

(a) kg s' 1 (b) kg s 

(c) kg m s 1 (d) kg m s' 2 

1.5 The number of significant digits in 0.0006032 is 
(a) 8 (b) 7 

(c) 4 (d) 2 

1.6 The length of a body is measured as 3.51 m. If the accuracy is 
0.01 m, then the percentage error in the measurement is 

(a) 351 % (b) 1 % 

(c) 0.28 % (d) 0.035 % 

1.7 The dimensional formula for gravitational constant is 
(a) A^I/V 2 (b) M^L 3 T 2 

(c) M^L 3 T 2 (d) M l L 3 T 2 



33 



1.8 The velocity of a body is expressed as v = (x/t) + yt. The dimensional 
formula for x is 

(a) ML°T° (b) M°LT° 

(c) M°L°T (d) MLT° 

1.9 The dimensional formula for Planck's constant is 
(a) MLT (b) ML 3 T 2 

(c) ML°T 4 (d) ML 2 T l 

1.10 have the same dimensional formula 

(a) Force and momentum (b) Stress and strain 

(c) Density and linear density (d) Work and potential energy 

1.11 What is the role of Physics in technology? 

1.12 Write a note on the basic forces in nature. 

1.13 Distinguish between fundamental units and derived units. 

1.14 Give the SI standard for (i) length (ii) mass and (Hi) time. 

1.15 Why SI system is considered superior to other systems? 

1.16 Give the rules and conventions followed while writing SI units. 

1.17 What is the need for measurement of physical quantities? 

1.18 You are given a wire and a metre scale. How will you estimate the 
diameter of the wire? 

1.19 Name four units to measure extremely small distances. 

1.20 What are random errors? How can we minimise these errors? 

1.21 Show that — gt 2 has the same dimensions of distance. 

1.22 What are the limitations of dimensional analysis? 

1.23 What are the uses of dimensional analysis? Explain with one 
example. 

Problems 

1.24 How many astronomical units are there in 1 metre? 



34 



1.25 If mass of an electron is 9.11 x 10 31 kg how many electrons would 
weigh 1 kg? 

1.26 In a submarine fitted with a SONAR, the time delay between generation 
of a signal and reception of its echo after reflection from an enemy 
ship is observed to be 73.0 seconds. If the speed of sound in water is 
1450 m s' 1 , then calculate the distance of the enemy ship. 

1.27 State the number of significant figures in the following: 

(i) 600900 (ii) 5212.0 (Hi) 6.320 (iv) 0.0631 (v)2.64xl0 24 

1.28 Find the value ofn 2 correct to significant figures, ifn = 3.14. 

1.29 5.74 g of a substance occupies a volume of 1.2 cm 3 . Calculate its 
density applying the principle of significant figures. 

1 .30 The length, breadth and thickness of a rectanglar plate are 4.234 m, 
1.005 m and 2.01 cm respectively. Find the total area and volume of 
the plate to correct significant figures. 

1 .31 The length of a rod is measured as 25.0 cm using a scale having an 
accuracy of 0. 1 cm. Determine the percentage error in length. 

1.32 Obtain by dimensional analysis an expressionfor the surface tension 
of a liquid rising in a capillary tube. Assume that the surface tension 
T depends on mass m of the liquid, pressure P of the liquid and 
radius r of the capillary tube (Take the constant k = ~ )■ 

1.33 The force F acting on a body moving in a circular path depends on 
mass m of the body, velocity v and radius r of the circular path. 
Obtain an expression for the force by dimensional analysis (Take 
the value ofk= 1). 

1.34 Check the correctness of the following equation by dimensinal 
analysis 

mi) 2 
(i) F = — 5— where F is force, m is mass, v is velocity and r is radius 
r 

(ii) n = ^~\l~r where n is f requency, g is acceleration due to gravity 

and I is length. 



35 



(iii) — tuv = mgh where m is mass, v is velocity, g is acceleration 

due to gravity and h is height. 
1 .35 Convert using dimensional analysis 

18 
(i) — kmph into m s 1 
5 

5 
(ii) — ms L into kmph 
18 

(iii) 13.6 g cm~ 3 into kg m~ 3 



Answers 

1.1 (a) 1.2 (a) 1.3 (b) 1.4 (a) 

1.5 (c) 1.6 (c) 1.7 (b) 1.8 (b) 

1.9 (d) 1.10 (d) 

1.24 6.68 x 10' 12 AU 1.25 1.097 x 10 30 

1.26 52.925 km 1.27 4, 5, 4, 3, 3 

1.28 9.86 1.29 4.8 gem: 3 

1.30 4.255 m 2 , 0.0855 m 3 1.31 0.4 % 



Pr 
1.32 T= — 1.33 F 



1.34 wrong, correct, wrong 

1.35 lms- 1 ,! kmph, 1.36 x 10 4 kg m~ 3 



mu 2 



36 



2. Kinematics 

Mechanics is one of the oldest branches of physics. It deals with 
the study of particles or bodies when they are at rest or in motion. 
Modern research and development in the spacecraft design, its automatic 
control, engine performance, electrical machines are highly dependent 
upon the basic principles of mechanics. Mechanics can be divided into 
statics and dynamics. 

Statics is the study of objects at rest; this requires the idea of 
forces in equilibrium. 

Dynamics is the study of moving objects. It comes from the Greek 
word dynamis which means power. Dynamics is further subdivided into 
kinematics and kinetics. 

Kinematics is the study of the relationship between displacement, 
velocity, acceleration and time of a given motion, without considering 
the forces that cause the motion. 

Kinetics deals with the relationship between the motion of bodies 
and forces acting on them. 

We shall now discuss the various fundamental definitions in 
kinematics. 

Particle 

A particle is ideally just a piece or a quantity of matter, having 
practically no linear dimensions but only a position. 

Rest and Motion 

When a body does not change its position with respect to time, then 
it is said to be at rest. 

Motion is the change of position of an object with respect to time. 
To study the motion of the object, one has to study the change in 
position (x,y,z coordinates) of the object with respect to the surroundings. 
It may be noted that the position of the object changes even due to the 
change in one, two or all the three coordinates of the position of the 

37 



objects with respect to time. Thus motion can be classified into three 
types : 

(i) Motion in one dimension 

Motion of an object is said to be one dimensional, if only one of 
the three coordinates specifying the position of the object changes with 
respect to time. Example : An ant moving in a straight line, running 
athlete, etc. 

(ii) Motion in two dimensions 

In this type, the motion is represented by any two of the three 
coordinates. Example : a body moving in a plane. 

(Hi) Motion in three dimensions 

Motion of a body is said to be three dimensional, if all the three 
coordinates of the position of the body change with respect to time. 

Examples : motion of a flying bird, motion of a kite in the sky, 
motion of a molecule, etc. 

2.1 Motion in one dimension (rectilinear motion) 

The motion along a straight line is known as rectilinear motion. 
The important parameters required to study the motion along a straight 
line are position, displacement, velocity, and acceleration. 

2.1.1 Position, displacement and distance travelled by the particle 

The motion of a particle can be described if its position is known 
continuously with respect to time. 

The total length of the path is the distance travelled by the particle 
and the shortest distance between the initial and final position of the 
particle is the displacement. 

The distance travelled by a i< - n< *' h 

particle, however, is different from its y ^ t ( 1 t ^ v 

displacement from the origin. For P 2 P t 

example, if the particle moves from a Fig 21 Distance and disp i a cement 
point O to position P : and then to 

38 



position P 2 , its displacement at the position P 2 is - x 2 from the origin 
but, the distance travelled by the particle is x 1 +x 1 +x 2 = (2x x +x 2 ) 
(Fig 2.1). 

The distance travelled is a scalar quantity and the displacement 
is a vector quantity. 

2.1.2 Speed and velocity 

Speed 

It is the distance travelled in unit time. It is a scalar quantity. 

Velocity 

The velocity of a particle is defined as the rate of change of 
displacement of the particle. It is also defined as the speed of the particle 
in a given direction. The velocity is a vector quantity. It has both 
magnitude and direction. 

displacement 

Velocity = — —. 

J time taken 

Its unit is m s _1 and its dimensional formula is LT -1 . 

Uniform velocity 

S A particle is said to move with uniform 

velocity if it moves along a fixed direction and 
covers equal displacements in equal intervals of 
time, however small these intervals of time may 
be. 

In a displacement - time graph, 
_^. £ (Fig. 2.2) the slope is constant at all the points, 



^ when the particle moves with uniform velocity. 

Fig. 2.2 Uniform velocity 

Non uniform or variable velocity 

The velocity is variable (non-uniform), if it covers unequal 
displacements in equal intervals of time or if the direction of motion 
changes or if both the rate of motion and the direction change. 



39 



Average velocity 

Let s 2 be the displacement of 
a body in time t x and s 2 be its 
displacement in time t 2 (Fig. 2.3). 
The average velocity during the time 
interval [t 2 - tj is defined as 

change in displacement 



v. 



average 



change in time 
>, As 



t 2 - 1, At 




► t 



From the graph, it is found 
that the slope of the curve varies. 



t, t 2 

Fig. 2.3 Average velocity 



Lt *L 

A£->0 At 



ds 
~dt 



Instantaneous velocity 

It is the velocity at any given instant of time or at any given point 
of its path. The instantaneous velocity v is given by 

v 

2.1.3 Acceleration 

If the magnitude or the direction or both of the velocity changes with 
respect to time, the particle is said to be under acceleration. 

Acceleration of a particle is defined as the rate of change of velocity. 

Acceleration is a vector quantity. 

change in velocity 

Acceleration = — 

time taken 

If u is the initial velocity and v, the final velocity of the particle 

after a time t, then the acceleration, 

v-u 

a ■- 



t 
Its unit is m s 2 and its dimensional formula is LT -2 . 

The instantaneous acceleration is, a = — = — — = — - 

dt dt{dt) dt 2 

Uniform acceleration 

If the velocity changes by an equal amount in equal intervals of 
time, however small these intervals of time may be, the acceleration is 
said to be uniform. 



40 



Retardation or deceleration 



If the velocity decreases with time, the acceleration is negative. The 
negative acceleration is called retardation or deceleration. 

Uniform motion 

A particle is in uniform motion when it moves with constant 
velocity (i.e) zero acceleration. 

2.1.4 Graphical representations 

The graphs provide a convenient method to present pictorially, 
the basic informations about a variety of events. Line graphs are used 
to show the relation of one quantity say displacement or velocity with 
another quantity such as time. 

If the displacement, velocity and acceleration of a particle are 
plotted with respect to time, they are known as, 

(i) displacement - time graph (s - t graph) 

(ii) velocity - time graph [v - t graph) 

(iii) acceleration - time graph (a - t graph) 

Displacement - time graph s 

When the displacement of the 
particle is plotted as a function of time, 
it is displacement - time graph. 



As v 



ds 
~dl 



the slope of the s - t 



graph at any instant gives the velocity 
of the particle at that instant. In 
Fig. 2.4 the particle at time t v has a O 
positive velocity, at time t 2 , has zero 
velocity and at time t 3 , has negative 
velocity. 




Fig. 2.4 Displacement 
time graph 



Velocity - time graph 

When the velocity of the particle is plotted as a function of time, 
it is velocity-time graph. 



dv 



As a = — , the slope of the v - t curve at any instant gives the 



dt 



41 



acceleration of the particle (Fig. 2.5). 

ds 
But, v = — - or as = v.dt 
at 

If the displacements are Sj and 
s 2 in times t 1 and t 2 , then 



f ds = [vdt 



k 




B 






i s . 




y V 




^ 


dt 






D 






C 


->t 



f 2 O t, h 

s 2~ s l = J v dt= area ABCD Fig 25 velocity - time graph 

ti 

The area under the u - t curve, between the given intervals of 
time, gives the change in displacement or the distance travelled by the 
particle during the same interval. a 

p 
Acceleration - time graph 



a dt 




Fig. 2.6 Acceleration 
- time graph 



When the acceleration is plotted as a 

function of time, it is acceleration - time 

graph (Fig. 2.6). 

dv 
a = — (or) dv = a dt 



If the velocities are v 1 and v 2 at times 
tj and <2 respectively, then 

J dv = J a dt (or) v 2 - Vj = ja.dt= area PQRS 
vi t! n 

The area under the a - t curve, between the given intervals of 
time, gives the change in velocity of the particle during the same interval. 
If the graph is parallel to the time axis, the body moves with constant 
acceleration. 

2.1.5 Equations of motion 

For uniformly accelerated motion, some simple equations that 
relate displacement s, time t, initial velocity u, final velocity v and 
acceleration a are obtained. 

(i) As acceleration of the body at any instant is given by the first 
derivative of the velocity with respect to time, 

42 



dv 
a = — (or) dv = a.dt 

If the velocity of the body changes from u to v in time t then from 
the above equation, 

v t t v t 

\dv = \adt = of dt => [v] u =a[t] Q 

u 

v - u = at 



(or) v = u + at 



• ••(I) 



(ii) The velocity of the body is given by the first derivative of the 
displacement with respect to time. 

ds 
(i.e) v = — (or) ds = v dt 

Since v = u + at, ds = (u + at) dt 
The distance s covered in time t is, 

t t 



jds = judt + jatdt ( or ) s = ut +—at 2 
ooo 2 



• ••(2) 



(iii) The acceleration is given by the first derivative of velocity with 
respect to time, (i.e) 

(or) ds = — v dv 
a 



dv dv 
dt ~ ds 


ds dv 
dt ds 


v v 


ds 
~~dl_ 


(( 


Therefore, 






s v j 
\ds=\ VdV 

o i a 


(i.e) s = 


1 
a 


2 2 
V U 

~2 ~2 



s = — (v 2 -u 2 ) (or) 2as = (v 2 - u 2 ) 

v 2 = u 2 + 2 as ...(3) 

The equations (1), (2) and (3) are called equations of motion. 

Expression for the distance travelled in n th second 

Let a body move with an initial velocity u and travel along a 
straight line with uniform acceleration a 

Distance travelled in the n th second of motion is, 

s n = distance travelled during first n seconds - distance 
travelled during (n -1) seconds 



43 



Distance travelled during n seconds 
D„ 



1 2 
un + —an 

2 



Distance travelled during (n -1) seconds 

1 



D 



(n-l) 



u(n-l) + y a(n-l) 2 



the distance travelled in the n th second = D- D 



1 2 
d.ei s„ =] un + —an 



r 



= u + a 



n 



\ 



(n-l) 



u(n-l) + -a(n-l) 2 
2 



s„ = u + —a(2n- 1) 
2 

Special Cases 

Case (i) : For downward motion 

For a particle moving downwards, a = g, since the particle moves 
in the direction of gravity. 

Case (ii) : For a freely falling body 

For a freely falling body, a = g and u = 0, since it starts from 
rest. 

Case (iii) : For upward motion 

For a particle moving upwards, a = - g, since the particle moves 
against the gravity. 

2.2 Scalar and vector quantities 

A study of motion will involve the introduction of a variety of 
quantities, which are used to describe the physical world. Examples 
of such quantities are distance, displacement, speed, velocity, 
acceleration, mass, momentum, energy, work, power etc. All these 
quantities can be divided into two categories - scalars and vectors. 

The scalar quantities have magnitude only. It is denoted by a 
number and unit. Examples : length, mass, time, speed, work, energy, 



44 



temperature etc. Scalars of the same kind can be added, subtracted, 
multiplied or divided by ordinary laws. 

The vector quantities have both magnitude and direction. Examples: 
displacement, velocity, acceleration, force, weight, momentum, etc. 



2.2.1 Representation of a vector 

Vector quantities are often represented by a scaled vector diagrams. 
Vector diagrams represent a vector by the use of an arrow drawn to 
scale in a specific direction. An example of a scaled vector diagram is 
shown in Fig 2.7. 

From the figure, it is clear that 

(i) The scale is listed. 

(ii) A line with an arrow is drawn in a specified direction. 

(iii) The magnitude and direction 

of the vector are clearly labelled. In 

the above case, the diagram shows that 

the magnitude is 4 N and direction is 

30° to x-axis. The length of the line 

gives the magnitude and arrow head 

gives the direction. In notation, the 

vector is denoted in bold face letter 

such as A or with an arrow above the 

-> 
letter as A, read as vector 

A or A vector while its magnitude O 

is denoted by A or by A . 



lcm=lN 




^Head 



2.2.2 Different types of vectors 



**->Tail 

Fig 2.7 Representation 
of a vector 



-> 
A 



B 
Fig. 2.8 
Equal vectors 



(i) Equal vectors 

Two vectors are said to be equal if they have the 

same magnitude and same direction, wherever be their 

-» — » 

initial positions. In Fig. 2.8 the vectors A and B have 

-> — » 

the same magnitude and direction. Therefore A and B 

are equal vectors. 



45 




A> 



B 



j? 



-> 
B 



Fig. 2.9 
Like uectors 



Fig. 2.10 
Opposite vectors 



Fig. 2.11 
Unlike Vectors 



(ii) Like vectors 

Two vectors are said to be like vectors, if they have same direction 
but different magnitudes as shown in Fig. 2.9. 

(Hi) Opposite vectors 

The vectors of same magnitude but opposite in direction, are 
called opposite vectors (Fig. 2.10). 

(iv) Unlike vectors 

The vectors of different magnitude acting in opposite directions 
are called unlike vectors. In Fig. 2.11 the vectors A and B are unlike 
vectors. 

(v) Unit vector 

A vector having unit magnitude is called a unit vector. It is also 

defined as a vector divided by its own magnitude. A unit vector in the 

-> A 

direction of a vector A is written as A and is read as A cap' or A caret' 

or A hat'. Therefore, 



a A 



(or) A = A |A| 



Thus, a vector can be written as the product of its magnitude and 
unit vector along its direction. 

Orthogonal unit vectors 

There are three most common unit vectors in the positive directions 
of X,Y and Z axes of Cartesian coordinate system, denoted by i, j and 
k respectively. Since they are along the mutually perpendicular directions, 
they are called orthogonal unit vectors. 

(vi) Null vector or zero vector 

A vector whose magnitude is zero, is called a null vector or zero 

-> 
vector. It is represented by and its starting and end points are the 

same. The direction of null vector is not known. 



46 



t 



(vii) Proper vector 

All the non-zero vectors are called proper vectors. 

(viii) Co-initial vectors 

Vectors having the same starting point are called 

— > — > 

co-initial vectors. In Fig. 2.12, A and B start from the Fic . 2 12 

same origin O. Hence, they are called as co-initial Co-initial vectors 

vectors. 

(ix) Coplanar vectors 

Vectors lying in the same plane are called coplanar vectors and 
the plane in which the vectors lie are called plane of vectors. 

2.2.3 Addition of vectors 

As vectors have both magnitude and direction they cannot be 
added by the method of ordinary algebra. 

Vectors can be added graphically or geometrically. We shall now 

discuss the addition of two vectors graphically using head to tail method. 

-> -» 

Consider two vectors P and Q which are acting along the same 

line. To add these two vectors, join the tail of Q with the head of P 
(Fig. 2.13). 

The resultant of P and Q is R = P + Q. The length of the line 
AD gives the magnitude of R. R acts in the same direction as that of 
P and Q. 



p q are inclined to each other, triangle law of vectors 

A BC D 
> > 



A B 



In order to find the sum of two vectors, which 
dined to each other, triangle law of vect 
or parallelogram law of vectors, can be used. 

-> 
C Q p (i) Triangle law of vectors 

If two vectors are represented in magnitude 
-^ and direction by the two adjacent sides of a triangle 

£ * D taken in order, then their resultant is the closing 

Fiq. 2.13 side of the triangle taken in the reverse order. 

Addition of vectors 



47 




Fig. 2.14 Triangle law of vectors 



To find the resultant of 
two vectors P and Q which 
are acting at an angle 9, the 
following procedure is adopted. 

First draw OA = p 

(Fig. 2.14) Then starting from 

-> 
the arrow head of P, draw the 

vector AB = Q ■ Finally, draw 

a vector OB = R from the 

tail of vector P to the head of vector Q. Vector OB = R is the sum 

of the vectors P and Q. Thus R = P + Q. 

-» -» 
The magnitude of P + Q is determined by measuring the length 
—> — > — > 

of R and direction by measuring the angle between P and R. 

— > 
The magnitude and direction of R, can be obtained by using the 

sine law and cosine law of triangles. Let a be the angle made by the 

resultant R with P. The magnitude of R is, 

R 2 = P 2 + Q 2 - 2PQ cos (180° - 0) 

R = ^P 2 +Q 2 +2PQ cosO 

The direction of R can be obtained by, 

P = 9 = R 

sin p sin a sin (180° -0) 

(ii) Parallelogram, law of vectors 

If two vectors acting at a point are represented in magnitude and 

direction by the two adjacent sides of a parallelogram, then their resultant 

is represented in magnitude and direction by the diagonal passing through 

the common tail of the two vectors. 

—> —> 

Let us consider two vectors P and Q which are inclined to each 

other at an angle 8 as shown in Fig. 2.15. Let the vectors P and Q be 

represented in magnitude and direction by the two sides OA and OB of 

a parallelogram OACB. The diagonal OC passing through the common 

tail O, gives the magnitude and direction of the resultant R. 

CD is drawn perpendicular to the extended OA, from C. Let 

| COD made by R with P be a. 



48 



From right angled triangle OCD, 
OC 2 = OD 2 + CD 2 

= (OA + AD) 2 + CD 2 

= OA 2 + AD 2 + 2.0A.AD + CD 2 



• (1) 




In Fig. 2.15 \BOA = 6 = |CAD 



...(2) 



Fig 2.15 Parallelogram 
law of vectors 



From right angled A CAD, 
AC 2 = AD 2 + CD 2 
Substituting (2) in (1) 
OC 2 = OA 2 + AC 2 + 20A.AD ...(3) 
From AACD, 

CD = AC sin ...(4) 

...(5) 



AD = AC cos 

Substituting (5) in (3) OC 2 = OA 2 + AC 2 + 2 OA.AC cos 6 

Substituting OC = R, OA = P, 
OB = AC = Q in the above equation 
R 2 = P 2 + Q 2 + 2PQ cos 6 



(or) r = J p 2 + 



(6) 



+ 2PQ cos 

Equation (6) gives the magnitude of the resultant. From A OCD, 

CD _ CD 

OD ~ 



tan a 



OA + AD 
Substituting (4) and (5) in the above equation, 

AC sin <9 Q sin 



tan a = 



(or) 



OA + AC cos P + Q cos 6 
Q sin B 



a = tan 



P + Q cos 
Equation (7) gives the direction of the resultant. 

Special Cases 

(i) When two vectors act in the same direction 

In this case, the angle between the two vectors 
cos 0° = 1, sin 0°= 



.(7) 



= 0°, 



49 



From (6) R = ^ P 2 + Q 2 + 2PQ = (P + Q) 



From (7) a = tan 1 



Q sin 0° 



P + Q cos 0° 

(i.e) a = 

Thus, the resultant vector acts in the same direction as the 
individual vectors and is equal to the sum of the magnitude of the two 
vectors. 

(ii) When two vectors act in the opposite direction 

In this case, the angle between the two vectors 8 = 180°, 
cos 180° = -1, sin 180° = 0. 



From (6) r ^ J p 2 + Q 2 - 1PQ --= (P - Q) 





From (7) a = tan _1 



p-9 



tan _1 (0) = 



Thus, the resultant vector has a magnitude equal to the difference 
in magnitude of the two vectors and acts in the direction of the bigger 
of the two vectors 

(Hi) When two vectors are at right angles to each other 
In this case, 6 = 90°, cos 90° = 0, sin 90° = 1 



2 



From (6) r = ^] P 2 + Q 

From (7) a = tan" 1 ^ 

The resultant R vector acts at an angle a with vector P. 

2.2.4 Subtraction of vectors 

The subtraction of a vector from another is equivalent to the 
addition of one vector to the negative of the other. 

For example Q* - P* = Q* + (-?) . 

-> -> -> -> 

Thus to subtract P from Q, one has to add - P with Q 

-> -> -> 

(Fig 2.16a). Therefore, to subtract P from Q, reversed P is added to the 

50 



Q . For this, first draw A B = Q and then starting from the arrow head 

-> — . _ -> -> 

of Q, draw B C = (-P ) and finally join the head of - P . Vector R is the 

—> -> -> -> 

sum of Q and - P. (i.e) difference Q - P. 



Q Y 






-> 
P 
A 


-> 

Q 

BCD 
— »• < 


A 9 B 






C 

4 


D 
i 


\ y 


A 


B 


C 






-> 
R 

A C 




(a) 








(b) 


Fig 


2.16 Subtraction 


of 


vectors 





The resultant of two vectors which are antiparallel to each other 
is obtained by subtracting the smaller vector from the bigger vector as 
shown in Fig 2.16b. The direction of the resultant vector is in the 
direction of the bigger vector. 

2.2.5 Product of a vector and a scalar 

Multiplication of a scalar and a vector gives a vector quantity 
which acts along the direction of the vector. 

Examples 

(i) If a is the acceleration produced by a particle of mass m under 

—> —> 

the influence of the force, then F = ma 

—> —> 
(ii) momentum = mass x velocity (i.e) P = mv. 

2.2.6 Resolution of vectors and rectangular components 

A vector directed at an angle with the co-ordinate axis, can be 
resolved into its components along the axes. This process of splitting a 
vector into its components is known as resolution of a vector. 

Consider a vector R = A making an angle 9 with X - axis. The 

vector R can be resolved into two components along X - axis and 
Y-axis respectively. Draw two perpendiculars from A to X and Y axes 
respectively. The intercepts on these axes are called the scalar 
components R x and R . 

51 



Then, OP Is R x , which is the magnitude of x component of R and 

— > 
OQ is R which is the magnitude of y component of R 




From A OP A, 




0P 


R„ 


cos 6 = — — = 
OA 


" R 


sin B = = 


,h 



(or) R r = R cos 6 



(or) R = R sin 

OA R y 



and R 2 = R. 2 + R. 



Fig. 2.17 Rectangular . y > 

components of a vector R = RJ + R j where i and j are unit vectors. 



Also, R can be expressed as 



In terms of R x and R , 9 can be expressed as B = tan J 



2.2.7 Multiplication of two vectors 

Multiplication of a vector by another vector does not follow the 
laws of ordinary algebra. There are two types of vector multiplication 
(i) Scalar product and (ii) Vector product. 

(i) Scalar product or Dot product of two 
vectors 

If the product of two vectors is a scalar, 
then it is called scalar product. If A and B are 
two vectors, then their scalar product is written B 

as A.B and read as A dot B. Hence scalar product Fia 2 _l 8 ^ Sc ; al " r _^™ duct 
is also called dot product. This is also referred as 
inner or direct product. 




of two vectors 



The scalar product of two vectors is a scalar, which is equal to 

the product of magnitudes of the two vectors and the cosine of the 

— > — > 

angle between them. The scalar product of two vectors A and B may 

— > — > — > — > — > — > 

be expressed as A . B = \ A \ \B\ cos 9 where | A | and | B \ are the 

magnitudes of A and B respectively and 8 is the angle between A and 

— > 

B as shown in Fig 2.18. 



52 



(ii) Vector product or Cross product of two vectors 

If the product of two vectors is a vector, then it is called vector 

-> -> 

product. If A and B are two vectors then their vector product is written 

as A x B and read as A cross B. This is also referred as outer product. 

The vector product or cross product of two vectors is a vector 
whose magnitude is equal to the product of their magnitudes and the 
sine of the smaller angle between them and the direction is perpendicular 
to a plane containing the two vectors. 



o 



AX B 



O 



BX A 



-> 




If is the smaller angle through which 
A should be rotated to reach B, then the cross 
product of A and B (Fig. 2.19} is expressed 
^ as, 



f 



A x B 



I B I sin 



A 

n 



C 



o 



Fig 2.19 Vector product 
of two vectors 



where \A \ and | B \ are the magnitudes of A 

and B respectively. C is perpendicular to the 

plane containing A and B. The direction of C 

is along the direction in which the tip of a 

screw moves when it is rotated from A to B. 

-> 
Hence C acts along OC. By the same 

-> -> 
argument, B x A acts along OD. 



2.3 Projectile motion 

A body thrown with some initial velocity and then allowed to move 
under the action of gravity alone, is known as a projectile. 

If we observe the path of the projectile, we find that the projectile 
moves in a path, which can be considered as a part of parabola. Such 
a motion is known as projectile motion. 

A few examples of projectiles are (i) a bomb thrown from an 
aeroplane (ii) a javelin or a shot-put thrown by an athlete (iii) motion 
of a ball hit by a cricket bat etc. 

The different types of projectiles are shown in Fig. 2.20. A body 
can be projected in two ways: 



53 




9 o 



Fig 2.20 Different types of projectiles 

(i) It can be projected horizontally from a certain height. 

(il) It can be thrown from the ground in a direction inclined 
to it. 

The projectiles undergo a vertical motion as well as horizontal 
motion. The two components of the projectile motion are (i) vertical 
component and (ii) horizontal component. These two perpendicular 
components of motion are independent of each other. 

A body projected with an initial velocity making an angle with the 
horizontal direction possess uniform horizontal velocity and variable 
vertical velocity, due to force of gravity. The object therefore has 
horizontal and vertical motions simultaneously. The resultant motion 
would be the vector sum of these two motions and the path following 
would be curvilinear. 

The above discussion can be summarised as in the Table 2.1 



Table 2 


.1 Two independent motions of a 


projectile 


Motion 


Forces 


Velocity 


Acceleration 


Horizontal 


No force acts 


Constant 


Zero 


Vertical 


The force of 


Changes 


Downwards 




gravity acts 


(~10 m s _1 ) 


(~10m s 2 ) 




downwards 







In the study of projectile motion, it is assumed that the air 
resistance is negligible and the acceleration due to gravity remains 
constant. 



54 



Angle of projection 

The angle between the initial direction of projection and the horizontal 
direction through the point of projection is called the angle of projection. 

Velocity of projection 

The velocity with which the body is projected is known as velocity 
of projection. 

Range 

Range of a projectile is the horizontal distance between the point of 
projection and the point where the projectile hits the ground. 

Trajectory 

The path described by the projectile is called the trajectory. 

Time of flight 

Time of flight is the total time taken by the projectile from the 
instant of projection till it strikes the ground. 

2.3.1 Motion of a projectile thrown horizontally 

Let us consider an object thrown horizontally with a velocity u 

from a point A, which is at a height 
h from the horizontal plane OX 
(Fig 2.21). The object acquires the 
following motions simultaneously : 

(i) Uniform velocity with which 
it is projected in the horizontal 
direction OX 

(ii) Vertical velocity, which is 
non-uniform due to acceleration due 
i» X to gravity. 

° M M 

R , , The two velocities are 

Fig 2.21 Projectile projected 
horizontally from the top of a tower independent of each other. The 

horizontal velocity of the object shall 
remain constant as no acceleration is acting in the horizontal direction. 
The velocity in the vertical direction shall go on changing because of 
acceleration due to gravity. 



55 




Path of a projectile 

Let the time taken by the object to reach C from A = t 
Vertical distance travelled by the object in time t = s = y 

From equation of motion, s = u^ + — at 2 ■■■[l] 

Substituting the known values in equation (1), 

y = (0) t + - gt 2 = -gt 2 ...(2) 

At A, the initial velocity in the horizontal direction is u. 

Horizontal distance travelled by the object in time t is x. 

x 
x = horizontal velocity x time = u t (or) t = — -.-(3) 

1 fxf 1 x 2 



Substituting t in equation (2), y = -g\ — \ = — g — ^ ...(4) 

(or) y = kx 2 

g 
where k = — o is a constant. 
2ir 

The above equation is the equation of a parabola. Thus the path 

taken by the projectile is a parabola. 

Resultant velocity at C 

At an instant of time t, let the body be at C 

At A, initial vertical velocity (Uj) = 

At C, the horizontal velocity (uj = u 

At C, the vertical velocity = u 2 

From equation of motion, u 2 = u 1 + g t 

Substituting all the known values, u 2 = + g t 

The resultant velocity at C is v = Juf + u. 2 , = Ju. 2 + g 2 t 

u 2 gt 
The direction of v is given by tan 8 = — = — ...(7) 

& J u x u 

where 8 is the angle made by v with X axis. 

56 



c 


u 




<] 1 > 

\e i 
\ i 


n 


\ i 

V 


Re 


f 

Fig 2.22 
sultant velocity 
at any point 


t ...(5) 


. 2 4 


q 2 t 2 ...(6) 



Time of flight and range 

The distance OB = R, is called as range of the projectile. 
Range = horizontal velocity x time taken to reach the ground 



R = u t 



:f 



...(8) 



where t f is the time of flight 

At A, initial vertical velocity (Uj) = 
The vertical distance travelled by the object in time t f = s 



From the equations of motion S 



"i*/ + oS*/ 



Substituting the known values in equation (9), 

l2h 



h = (0)tf + -gt} 



(or) t 



'J 



9 



Substituting t f in equation (8), Range R = u 



...(9) 

...(10) 
...(11) 



2.3.2 Motion of a projectile projected at an angle with the 
horizontal (oblique projection) 

Consider a body projected from a point O on the surface of the 
Earth with an initial velocity u at an angle 8 with the horizontal as 
shown in Fig. 2.23. The velocity u can be resolved into two components 



A (u,=0) 




u x =u cos I 

Fig 2.23 Motion of a projectile projected at an angle with horizontal 

57 



(i) u x = u cos , along the horizontal direction OX and 

(ii) u = u sin 6, along the vertical direction OY 

The horizontal velocity u x of the object shall remain constant as 
no acceleration is acting in the horizontal direction. But the vertical 
component u of the object continuously decreases due to the effect of 
the gravity and it becomes zero when the body is at the highest point 
of its path. After this, the vertical component u is directed downwards 
and increases with time till the body strikes the ground at B. 

Path of the projectile 

Let t 1 be the time taken by the projectile to reach the point C from 
the instant of projection. 

Horizontal distance travelled by the projectile in time t 1 is, 

x = horizontal velocity x time 

x 

x = u cos 8 x t, (or) t, = — ...(1) 

J 1 u cos 6 

Let the vertical distance travelled by the projectile in time 

tj = s = y 
At O, initial vertical velocity u T = u sin 8 

From the equation of motion s = u 1 t 1 - —gt? 
Substituting the known values, 
y = (u sin 9) tj - ± gt ? ...(2) 

Substituting equation (1) in equation (2), 

y = (u sin 9) { ^— 1 - \(g) ( X 



u cos Q ) 2 \u cos 8 

ax 2 
y=xtanQ - y ...(3) 

2u cos 8 

The above equation is of the form y = Ax + Bx 2 and represents 
a parabola. Thus the path of a projectile is a parabola. 

Resultant velocity of the projectile at any instant t x 

At C, the velocity along the horizontal direction is u x = u cos 8 and 
the velocity along the vertical direction is u = u 2 . 

58 



From the equation of motion, 

u 2 = u sin 8 - gt 1 
:. The resultant velocity at 



i£ 



C is v = ^u 2 x + 
v = yj[u cos 6) 2 + (u sin 9 - gt x f 




Fig 2.24 Resultant velocity of the 
projectile at any instant 



= ^u 2 + g 2 tf - 2ut x g sin 9 
The direction of v is given by 



tan or 



U2 u sin 8 - gt x 
u u cos 8 



(or) a = tan 



u sin 8 - gt Y 
u cos 8 



where a is the angle made by v with the horizontal line. 

Maximum height reached by the projectile 

The maximum vertical displacement produced by the projectile is 
known as the maximum height reached by the projectile. In Fig 2.23, 
EA is the maximum height attained by the projectile. It is represented 

as "mnr 

At O, the initial vertical velocity [uj = u sin 8 
At A, the final vertical velocity (u 3 ) = 

The vertical distance travelled by the object = s = h max 
From equation of motion, u| = uj - 2gs 
Substituting the known values, (0) 2 = (u sinQ) 2 - 2gh max 

u 2 sin 2 6 



2 9 h max = u sin 9 (° r ) 



2g 



.(4) 



Time taken to attain maximum height 

Let t be the time taken by the projectile to attain its maximum 
height. 

From equation of motion u 3 = u 1 - g t 



59 



Substituting the known values = u sin - g t 

g t = u sin 8 

u sin 6 
* = — r -.(5) 

Time of flight 

Let tr be the time of flight (i.e) the time taken by the projectile to 
reach B from O through A. When the body returns to the ground, the 
net vertical displacement made by the projectile 

s = h - h =0 

y max max 



From the equation of motion s = u 1 t f - —gtf 

1 , 2 
Substituting the known values = ( u sin Q) t f - ~9 £/ 



1 2 2usin 8 

-9 t f = (u sin 6} tj (or) t f = " ...(6) 

From equations (5) and (6) t f = 2t ■■■(7) 

(i.e) the time of flight is twice the time taken to attain the maximum 
height. 

Horizontal range 

The horizontal distance OB is called the range of the projectile. 
Horizontal range = horizontal velocity x time of flight 

(i.e) R = u cos 8 x t. 

_, 2usin0 
Substituting the value of t, R - (u cose) 



R = 



u 2 (2 sine cos 9) 



u 2 sin 26 



.". R = ...(8) 

9 

Maximum Range 

From (8), it is seen that for the given velocity of projection, the 
horizontal range depends on the angle of projection only. The range is 
maximum only if the value of sin 28 is maximum. 

60 



For maximum range R max sin 28 = 1 

(i.e) 6 = 45° 

Therefore the range is maximum when the angle of projection 
is 45°. 

R max ~ " => R max = ~^~ •••f 9 ' 

2.4 Newton's laws of motion 

Various philosophers studied the basic ideas of cause of motion. 
According to Aristotle, a constant external force must be applied 
continuously to an object in order to keep it moving with uniform 
velocity. Later this idea was discarded and Galileo gave another idea on 
the basis of the experiments on an inclined plane. According to him, no 
force is required to keep an object moving with constant velocity. It is 
the presence of frictional force that tends to stop moving object, the 
smaller the frictional force between the object and the surface on which 
it is moving, the larger the distance it will travel before coming to rest. 
After Galileo, it was Newton who made a systematic study of motion 
and extended the ideas of Galileo. 

Newton formulated the laws concerning the motion of the object. 
There are three laws of motion. A deep analysis of these laws lead us 
to the conclusion that these laws completely define the force. The first 
law gives the fundamental definition of force; the second law gives the 
quantitative and dimensional definition of force while the third law 
explains the nature of the force. 

2.4.1 Newton's first law of motion 

It states that every body continues in its state of rest or of uniform 
motion along a straight line unless it is compelled by an external force to 
change that state. 

This law is based on Galileo's law of inertia. Newton's first law of 
motion deals with the basic property of matter called inertia and the 
definition of force. 

Inertia is that property of a body by virtue of which the body is 
unable to change its state by itself in the absence of external force. 

61 



The inertia is of three types 
(i) Inertia of rest 
(ii) Inertia of motion 
(iii) Inertia of direction. 

(i) Inertia of rest 

It is the inability of the body to change its state of rest by itself. 

Examples 

(i) A person standing in a bus falls backward when the bus 
suddenly starts moving. This is because, the person who is initially at 
rest continues to be at rest even after the bus has started moving. 

(ii) A book lying on the table will remain at rest, until it is moved 
by some external agencies. 

(iii) When a carpet is beaten by a stick, the dust particles fall off 
vertically downwards once they are released and do not move along the 
carpet and fall off. 

(ii) Inertia of motion 

Inertia of motion is the inability of the body to change its state of 
motion by itself. 

Examples 

(a) When a passenger gets down from a moving bus, he falls down 
in the direction of the motion of the bus. 

(b) A passenger sitting in a moving car falls forward, when the car 
stops suddenly. 

(c) An athlete running in a race will continue to run even after 
reaching the finishing point. 

(iii) Inertia of direction 

It is the inability of the body to change its direction of motion by 
itself. 

Examples 

When a bus moving along a straight line takes a turn to the right, 
the passengers are thrown towards left. This is due to inertia which 
makes the passengers travel along the same straight line, even though 
the bus has turned towards the right. 

62 



This inability of a body to change by itself its state of rest or of 
uniform motion along a straight line or direction, is known as inertia. The 
inertia of a body is directly proportional to the mass of the body. 

From the first law, we infer that to change the state of rest or 
uniform motion, an external agency called, the force is required. 

Force is defined as that which when acting on a body changes or 
tends to change the state of rest or of uniform motion of the body along 
a straight line. 

A force is a push or pull upon an object, resulting the change of 
state of a body. Whenever there is an interaction between two objects, 
there is a force acting on each other. When the interaction ceases, the 
two objects no longer experience a force. Forces exist only as a result 
of an interaction. 

There are two broad categories of forces between the objects, 
contact forces and non-contact forces resulting from action at a distance. 

Contact forces are forces in which the two interacting objects are 
physically in contact with each other. 

Tensional force, normal force, force due to air resistance, applied 
forces and frictional forces are examples of contact forces. 

Action-at-a-distance forces (non- contact forces) are forces in which 
the two interacting objects are not in physical contact which each other, 
but are able to exert a push or pull despite the physical separation. 
Gravitational force, electrical force and magnetic force are examples of 
non- contact forces. 

Momentum of a body 

It is observed experimentally that the force required to stop a 
moving object depends on two factors: (i) mass of the body and 
(ii) its velocity 

A body in motion has momentum. The momentum of a body is 
defined as the product of its mass and velocity. If m is the mass of the 
body and v, its velocity, the linear momentum of the body is given by 
p = mv. 

Momentum has both magnitude and direction and it is, therefore, 
a vector quantity. The momentum is measured in terms of kg m s 1 
and its dimensional formula is MLT -1 . 



63 



When a force acts on a body, its velocity changes, consequently, 
its momentum also changes. The slowly moving bodies have smaller 
momentum than fast moving bodies of same mass. 

If two bodies of unequal masses and velocities have same 
momentum, then, 



(i.e) 



Pi = P 2 

_> m t = v 2 

m 2 VI 



Hence for bodies of same momenta, their velocities are inversely 
proportional to their masses. 

2.4.2 Newton's second law of motion 

Newton's first law of motion deals with the behaviour of objects 
on which all existing forces are balanced. Also, it is clear from the first 
law of motion that a body in motion needs a force to change the 
direction of motion or the magnitude of velocity or both. This implies 
that force is such a physical quantity that causes or tends to cause an 
acceleration. 

Newton's second law of motion deals with the behaviour of objects 
on which all existing forces are not balanced. 

According to this law, the rate of change of momentum of a body 
is directly proportional to the external force applied on it and the change 
in momentum takes place in the direction of the force. 

If p is the momentum of a body and F the external force acting 
on it, then according to Newton's second law of motion, 

F a — — (or) F = k —— where k is a proportionality constant. 

If a body of mass m is moving with a velocity v then, its momentum 
is given by p = m v. 

— d — dv 

:. F = k — (m v) = k m — 
at dt 

Unit of force is chosen in such a manner that the constant k is 
equal to unity, (i.e) k =1. 

64 



.-. F 



m 



dv 
~dt 



— — > dv 

ma where a = —— is the acceleration produced 



in the motion of the body. 

The force acting on a body is measured by the product of mass of 
the body and acceleration produced by the force acting on the body. The 
second law of motion gives us a measure of the force. 

The acceleration produced in the body depends upon the inertia 
of the body (i.e) greater the inertia, lesser the acceleration. One newton 
is defined as that force which, when acting on unit mass produces unit 
acceleration. Force is a vector quantity. The unit of force is kg m s 2 or 
newton. Its dimensional formula is MLT . 



Impulsive force and Impulse of a force 
(i) Impulsive Force 

An impulsive force is a very great force acting for a very short time 
on a body, so that the change in the position of the body during the time 
the force acts on it may be neglected. 

(e.g.) The blow of a hammer, the collision of two billiard balls etc. 
(ii) Impulse of a force 

The impulse J of a constant force F F^ 
acting for a time t is defined as the product 
of the force and time. 

(i.e) Impulse = Force x time 

J = F x t 

The impulse of force F acting over a 

time interval t is defined by the integral, 
t 
J = j> dt 



.(1) 




Fig .2.25 Impulse of a force 



The impulse of a force, therefore can 
be visualised as the area under the force 

versus time graph as shown in Fig. 2.25. When a variable force acting 
for a short interval of time, then the impulse can be measured as, 



J = F. 



average 



x dt 



• (2) 



65 



Impulse of a force is a vector quantity and its unit is N s. 

Principle of impulse and momentum 

By Newton's second law of motion, the force acting on a 
body = ma where m = mass of the body and a = acceleration produced 

The impulse of the force = F x t = (m a) t 

If u and v be the initial and final velocities of the body then, 

(v-u) 
a = 

t ' 

Therefore, impulse of the force = mx — - — x t = m[v - u) = mv - mu 

Impulse = final momentum of the body 

- initial momentum of the body. 

(i.e) Impulse of the force = Change in momentum 

The above equation shows that the total change in the momentum 
of a body during a time interval is equal to the impulse of the force acting 
during the same interval of time. This is called principle of impulse and 
momentum. 

Examples 

(i) A cricket player while catching a ball lowers his hands in the 
direction of the ball. 

If the total change in momentum is brought about in a very 

short interval of time, the average force is very large according to the 

mv - mu 
equation, F = 

By increasing the time interval, the average force is decreased. It 
is for this reason that a cricket player while catching a ball, to increase 
the time of contact, the player should lower his hand in the direction 
of the ball , so that he is not hurt. 

(ii) A person falling on a cemented floor gets injured more where 
as a person falling on a sand floor does not get hurt. For the same 
reason, in wrestling, high jump etc., soft ground is provided. 

(iii) The vehicles are fitted with springs and shock absorbers to 
reduce jerks while moving on uneven or wavy roads. 

66 



2.4.3 Newton's third Law of motion 

It is a common observation that when we sit on a chair, our body 
exerts a downward force on the chair and the chair exerts an upward 
force on our body. There are two forces resulting from this interaction: 
a force on the chair and a force on our body. These two forces are 
called action and reaction forces. Newton's third law explains the relation 
between these action forces. It states that for every action, there is an 
equal and opposite reaction. 

(i.e.) whenever one body exerts a certain force on a second body, 
the second body exerts an equal and opposite force on the first. Newton's 
third law is sometimes called as the law of action and reaction. 

Let there be two bodies 1 and 2 exerting forces on each other. Let 
the force exerted on the body 1 by the body 2 be F 12 and the force 
exerted on the body 2 by the body 1 be F 21 . Then according to third 
law, F 12 = -F 21 . 

One of these forces, say F 12 may be called as the action whereas 
the other force F 21 may be called as the reaction or vice versa. This 
implies that we cannot say which is the cause (action) or which is the 
effect (reaction). It is to be noted that always the action and reaction 
do not act on the same body; they always act on different bodies. The 
action and reaction never cancel each other and the forces always exist 
in pair. 

The effect of third law of motion can be observed in many activities 
in our everyday life. The examples are 

(i) When a bullet is fired from a gun with a certain force (action), 
there is an equal and opposite force exerted on the gun in the backward 
direction (reaction). 

(ii) When a man jumps from a boat to the shore, the boat moves 
away from him. The force he exerts on the boat (action) is responsible 
for its motion and his motion to the shore is due to the force of reaction 
exerted by the boat on him. 

(iii) The swimmer pushes the water in the backward direction 
with a certain force (action) and the water pushes the swimmer in the 
forward direction with an equal and opposite force (reaction). 

67 



(iv) We will not be able to walk if there 
were no reaction force. In order to walk, we 
push our foot against the ground. The Earth 
in turn exerts an equal and opposite force. 
This force is inclined to the surface of the 
Earth. The vertical component of this force 
balances our weight and the horizontal 
component enables us to walk forward. 

(v) A bird flies by with the help of its 
wings. The wings of a bird push air downwards 



YA 



Reaction 




V- Action 

Fig. 2.25a Action and 



reaction 
(action). In turn, the air reacts by pushing the bird upwards (reaction). 

(vi) When a force exerted directly on the wall by pushing the palm 
of our hand against it (action), the palm is distorted a little because, 
the wall exerts an equal force on the hand (reaction). 

Law of conservation of momentum 

From the principle of impulse and momentum, 
impulse of a force, J = mv - mu 
If J = then mv-mu =0 (or) mu = mu 
(i.e) final momentum = initial momentum 

In general, the total momentum of the system is always a constant 
(i.e) when the impulse due to external forces is zero, the momentum of the 
system remains constant. This is known as law of conservation of 
momentum. 

We can prove this law, in the case of a head on collision between 
two bodies. 

Proof 

Consider a body A of mass m 1 moving with a velocity u 1 collides 
head on with another body B of mass m 2 moving in the same direction 
as A with velocity u 2 as shown in Fig 2.26. 





C^&* 



Before Collision During Collision After Collision 

Fig. 2. 26 Law of conservation of momentum 



68 



After collision, let the velocities of the bodies be changed to v 1 
and v 2 respectively, and both moves in the same direction. During 
collision, each body experiences a force. 

The force acting on one body is equal in magnitude and opposite 
in direction to the force acting on the other body. Both forces act for 
the same interval of time. 

Let F 1 be force exerted by A on B (action), F 2 be force exerted by 
B on A (reaction) and t be the time of contact of the two bodies during 
collision. 

Now, F 1 acting on the body B for a time t, changes its velocity 
from u 2 to v 2 . 

:. F 1 = mass of the body B x acceleration of the body B 

= m 9 x i-2 *L ...(1) 

2 t 

Similarly, F 2 acting on the body A for the same time t changes its 

velocity from u 1 to v, 

.'. F 2 = mass of the body A x acceleration of the body A 
(«i - "J 



(i.e) m„ x 



-i t 

Then by Newton's third law of motion F, = -F„ 
(v 2 - u 2 ) (t)j - u x ) 



.-(2) 



2 t ~ "-i t 

m 2 (v 2 - u 2 ) = - m 1 (v 1 - uj 

m 2 v 2 - m 2 u 2 = - m 1 v 1 + m 1 u 1 

m 1 Uj + m 2 u 2 = m 1 v x + m 2 v 2 ...(3) 

(i.e) total momentum before impact = total momentum after impact, 
(i.e) total momentum of the system is a constant. 

This proves the law of conservation of linear momentum. 

Applications of law of conservation of momentum 

The following examples illustrate the law of conservation of 
momentum. 

(i) Recoil of a gun 

Consider a gun and bullet of mass m and m fa respectively. The 
gun and the bullet form a single system. Before the gun is fired, both 

69 



the gun and the bullet are at rest. Therefore the velocities of the gun 
and bullet are zero. Hence total momentum of the system before firing 
is m g (0) + m b (0) = 

When the gun is fired, the bullet moves forward and the gun 
recoils backward. Let v b and v are their respective velocities, the total 
momentum of the bullet - gun system, after firing is m b v b + rnv 

According to the law of conservation of momentum, total 
momentum before firing is equal to total momentum after firing. 



(i.e) = m. d.+ m u (or) v n 



m g 



It is clear from this equation, that v is directed opposite to v b . 
Knowing the values of m b , m and v b , the recoil velocity of the gun v 
can be calculated. 

(ii) Explosion of a bomb 

Suppose a bomb is at rest before it explodes. Its momentum is 
zero. When it explodes, it breaks up into many parts, each part having 
a particular momentum. A part flying in one direction with a certain 
momentum, there is another part moving in the opposite direction with 
the same momentum. If the bomb explodes into two equal parts, they 
will fly off in exactly opposite directions with the same speed, since 
each part has the same mass. 

Applications of Newton's third law of motion 

(i) Apparent loss of weight in a lift 

Let us consider a man of mass M standing on a weighing machine 
placed inside a lift. The actual weight of the man = Mg. This weight 
(action) is measured by the weighing machine and in turn, the machine 
offers a reaction R. This reaction offered by the surface of contact on 
the man is the apparent weight of the man. 

Case (i) 

When the lift is at rest: 
The acceleration of the man = 
Therefore, net force acting on the man = 
From Fig. 2.27(i), R - Mg = (or) R = Mg 

70 




a=0 




a 



R 



L 



a 



Mg Mg 

(i) (W m 

Fig 2.27 Apparent loss of weight in a lift 

That is, the apparent weight of the man is equal to the actual 
weight. 

Case (ii) 

When the lift is moving uniformly in the upward or downward 
direction: 

For uniform motion, the acceleration of the man is zero. Hence, 
in this case also the apparent weight of the man is equal to the actual 
weight. 

Case (iii) 

When the lift is accelerating upwards: 

If a be the upward acceleration of the man in the lift, then the 
net upward force on the man is F = Ma 

From Fig 2.27(ii}, the net force 

F = R - Mg = Ma (or) R = M ( g + a ) 

Therefore, apparent weight of the man is greater than actual 
weight. 

Case (iv) 

When the lift is accelerating downwards: 

Let a be the downward acceleration of the man in the lift, then 
the net downward force on the man is F = Ma 
From Fig. 2.27 (iii), the net force 
F = Mg - R = Ma (or) R = M (g - a) 



71 



Therefore, apparent weight of the man is less than the actual 
weight. 

When the downward acceleration of the man is equal to the 
acceleration due to the gravity of earth, (i.e) a = g 

:. R = M (g - g) = 

Hence, the apparent weight of the man becomes zero. This is 
known as the weightlessness of the body. 

(ii) Working of a rocket and jet plane 

The propulsion of a rocket is one of the most interesting examples 
of Newton's third law of motion and the law of conservation of momentum. 
The rocket is a system whose mass varies with time. In a rocket, 
the gases at high temperature and pressure, produced by the 
combustion of the fuel, are ejected from a nozzle. The reaction of the 
escaping gases provides the necessary thrust for the launching and 
flight of the rocket. 

From the law of conservation of linear momentum, the momentum 
of the escaping gases must be equal to the momentum gained by the 
rocket. Consequently, the rocket is propelled in the forward direction 
opposite to the direction of the jet of escaping gases. Due to the thrust 
imparted to the rocket, its velocity and acceleration will keep on 
increasing. The mass of the rocket and the fuel system keeps on 
decreasing due to the escaping mass of gases. 

2.5 Concurrent forces and Coplanar forces 

The basic knowledge of various kinds 
of forces and motion is highly desirable for 
engineering and practical applications. The 
Newton's laws of motion defines and gives 
the expression for the force. Force is a vector 
quantity and can be combined according to 
the rules of vector algebra. A force can be 
graphically represented by a straight line with 
an arrow, in which the length of the line is 
proportional to the magnitude of the force and the arrowhead indicates 
its direction. 




*F 5 
Fig 2.28 Concurrent forces 



72 



F 



F, 



A force system is said to be 
concurrent, if the lines of all forces intersect 
at a common point (Fig 2.28). 

A force system is said to be coplanar, 

if the lines of the action of all forces lie in 

nv n on ^ i f one plane (Fig 2.29). 

Fig 2.29. Coplanar forces c & 

2.5.1 Resultant of a system of forces acting on a rigid body 

If two or more forces act simultaneously on a rigid body, it is 
possible to replace the forces by a single force, which will produce the 
same effect on the rigid body as the effect produced jointly by several 
forces. This single force is the resultant of the system of forces. 

-» -> 

If P and Q are two forces acting on a body simultaneously in the 

same direction, their resultant is R = P + Q and it acts in the same 

direction as that of the forces. If P and Q act in opposite directions, 

their resultant R is R = P ~ Q and the resultant is in the direction of 

the greater force. 

-> -> 

If the forces P and Q act in directions which are inclined to each 

other, their resultant can be found by using parallelogram law of forces 

and triangle law of forces. 

2.5.2 Parallelogram law of forces 

If two forces acting at a point are represented 
in magnitude and direction by the two adjacent 
sides of a parallelogram, then their resultant is 
represented in magnitude and direction by the 
diagonal passing through the point. O 

Explanation b 

Consider two forces P and Q 

acting at a point O inclined at an angle 

8 as shown in Fig. 2.30. 

-> -> 

The forces P and Q are 

represented in magnitude and 

direction by the sides OA and OB of 

a parallelogram OACB as shown in 

Fig 2.30. 

73 




Fig 2.30 Parallelogram 
law of forces 



The resultant R of the forces P and Q is the diagonal OC of the 
parallelogram. The magnitude of the resultant is 

R = {p~ 2 



£T + 2PQ cos9 
The direction of the resultant is a 



tan 



Q sin 9 
P+Q cos 9 



2.5.3 Triangle law of forces 

The resultant of two forces acting at a point can also be found by 
using triangle law of forces. 

If two forces acting at a point 

are represented in magnitude and 

direction by the two adjacent sides 

of a triangle taken in order, then the 

closing side of the triangle taken in 

the reversed order represents the 

resultant of the forces in magnitude 

and direction. 

-> — » 

Forces P and Q act at an 

angle 8. In order to find the 

resultant of P and Q, one can apply 

the head to tail method, to construct 

the triangle. 




Fig 2.31 Triangle law of forces 



In Fig. 2.31, OA and AB represent P and Q in magnitude and 
direction. The closing side OB of the triangle taken in the reversed 
order represents the resultant R of the forces P and Q. The magnitude 
and the direction of R can be found by using sine and cosine laws of 
triangles. 

The triangle law of forces can also be stated as, if a body is in 
equilibrium under the action of three forces acting at a point, then the 
three forces can be completely represented by the three sides of a triangle 
taken in order. 

If P , Q and R are the three forces acting at a point and they 

P Q R 



are represented by the three sides of a triangle then 



OA AB OB 



74 



2.5.4 Equilibrant 

According to Newton's second law of motion, a body moves with 
a velocity if it is acted upon by a force. When the body is subjected to 
number of concurrent forces, it moves in a direction of the resultant 
force. However, if another force, which is equal in magnitude of the 
resultant but opposite in direction, is applied to a body, the body comes 
to rest. Hence, equilibrant of a system of forces is a single force, which 
acts along with the other forces to keep the body in equilibrium. 

Let us consider the forces F r F 2 , F 3 and F 4 acting on a body O 
as shown in Fig. 2.32a. If F is the resultant of all the forces and in 
order to keep the body at rest, an equal force (known as equilibrant) 
should act on it in the opposite direction as shown in Fig. 2.32b. 





> F 



*0y / 



(a) (b) 

Fig 2.32 Resultant and equilibrant 

From Fig. 2.32b, it is found that, resultant = - equilibrant 

2.5.5 Resultant of concurrent forces 

Consider a body O, which is acted upon by four forces as shown 
in Fig. 2.33a. Let 8 1 , 8 2 , 8 3 and 8 4 be the angles made by the forces with 
respect to X-axis. 

Each force acting at O can be replaced by its rectangular 
components F lx and F., F 2x and F 2 .. etc., 

For example, for the force Fj making an angle 8 p its components 
are, F lx =F 1 cos 1 and F 1 = F 1 sin 1 

These components of forces produce the same effect on the body 
as the forces themselves. The algebraic sum of the horizontal components 



75 




>x 




(b) 
Fig 2.33 Resultant of several concurrent forces 

F lx , F 2x , F 3x , .. gives a single horizontal component R 

(i- e ) R X = F 1 X + F 2 X + F 3 X + F 4 X = ^ F X 

Similarly, the algebraic sum of the vertical components F l , F 2 
F 3 , .. gives a single vertical component R 

(i.e) R y =F ly + F 2y + F 3y + F 4y = ZF y 

Now, these two perpendicular components R x and R can be added 
vectorially to give the resultant R . 



From Fig. 2.33b, R 



R x + R y 



(or) 



and 



tan a = 



R, 



(or) a = tan 



-l 



R = 



K + R y 



R 



v 



R. 



2.5.6 Lami's theorem 

It gives the conditions of equilibrium for three forces acting at a 
point. Lami's theorem states that if three forces acting at a point are in 
equilibrium, then each of the force is directly proportional to the sine of 
the angle between the remaining two forces. 

Let us consider three forces P, Q and R acting at a point O 
(Fig 2.34). Under the action of three forces, the point O is at rest, then 
by Lami's theorem, 



76 



P cc sin a 

Q cc sin p 

and R cc sin y, then 

p 9 R 



sin a sin /? sin y 



constant 




2.5-7 Experimental verification of triangle law, 
parallelogram law and Lami's theorem 

Two smooth small pulleys are fixed, one each Lami's theorem 
at the top corners of a drawing board kept vertically 
on a wall as shown in Fig. 2.35. The pulleys should move freely 
without any friction. A light string is made to pass over both the 
pulleys. Two slotted weights P and Q (of the order of 50 g) are taken 
and are tied to the two free ends of the string. Another short string 
is tied to the centre of the first string at O. A third slotted weight R is 
attached to the free end of the short string. The weights P, Q and R are 
adjusted such that the system is at rest. 

C 





D 



R 



Fig 2.35 Lami's theorem - experimental proof 

The point O is in equilibrium under the action of the three forces 
P, Q and R acting along the strings. Now, a sheet of white paper is held 
just behind the string without touching them. The common knot O and 
the directions of OA, OB and OD are marked to represent in magnitude, 
the three forces P, Q and R on any convenient scale (like 50 g = 1 cm). 



77 



The experiment is repeated for different values of P, Q and R and the 
values are tabulated. 



To verify parallelogram law 

To determine the resultant of two forces P and Q, a parallelogram 
OACB is completed, taking OA representing P, OB representing Q and 
the diagonal OC gives the resultant. The length of the diagonal OC and 
the angle \COD are measured and tabulated (Table 2.2). 

OC is the resultant R of P and Q. Since O is at rest, this 
resultant R must be equal to the third force R (equilibrant) which acts 
in the opposite direction. OC = OD. Also, both OC and OD are acting 
in the opposite direction. ZCOD must be equal to 180°. 

If OC = OD and ZCOD = 180°, one can say that parallelogram 
law of force is verified experimentally. 

Table 2.2 Verification of parallelogram law 



S.No. 


P 


9 


R 


OA 


OB 


OD 
(R) 


OC 

(R 1 ) 


ZCOD 


1. 
2. 
3. 



















To verify Triangle Law 

According to triangle law of forces, the resultant of P (= OA = BC) 

and Q (OB) is represented in magnitude and direction by OC which is 

taken in the reverse direction. 

p g 
Alternatively, to verify the triangle law offerees, the ratios — — . — — 

and — — are calculated and are tabulated (Table 2.3). It will be found out 

that, all the three ratios are equal, which proves the triangle law of 

forces experimentally. 

Table 2.3 Verification of triangle law 



S.No. 


P 


9 


R 1 


OA 


OB 


OC 


P 
OA 


9 

OB 


R' 
OC 


1. 

2. 
3. 





















78 



To verify Lami's theorem 

To verify Lami's theorem, the angles between the three forces, P, 
Q and R (i.e) ZBOD= a, ZAOD= P and ZAOB= J are measured using 

P 



protractor and tabulated (Table 2.4). The ratios — 



9 R 

and 



sina ' sin/3 sln^ 

are calculated and it is found that all the three ratios are equal and 
this verifies the Lami's theorem. 



Table 2.4 Verification of Lami's theorem 



S.No. 


P 


9 


R 


a 


P 


Y 


P 
sina 


9 

sinp 


R 

siny 


1. 

2. 
3. 





















2.5.8 Conditions of equilibrium of a rigid body acted upon by a 
system of concurrent forces in plane 

(i) If an object is in equilibrium under the action of three forces, the 
resultant of two forces must be equal and opposite to the third force. 
Thus, the line of action of the third force must pass through the point of 
intersection of the lines of action of the other two forces. In other words, 
the system of three coplanar forces in equilibrium, must obey parallelogram 
law, triangle law of forces and Lami's theorem. This condition ensures 
the absence of translational motion in the system. 

(ii) The algebraic sum of the moments about any point must be 
equal to zero. 2 M = (i.e) the sum of clockwise moments about any 
point must be equal to the sum of anticlockwise moments about the 
same point. This condition ensures, the absence of rotational motion. 

2.6 Uniform circular motion 

The revolution of the Earth around the Sun, rotating fly wheel, 
electrons revolving around the nucleus, spinning top, the motion of a 
fan blade, revolution of the moon around the Earth etc. are some 
examples of circular motion. In all the above cases, the bodies or 
particles travel in a circular path. So, it is necessary to understand the 
motion of such bodies. 

79 



When a particle moves on a circular 
path with a constant speed, then its 
motion is known as uniform circular 
motion in a plane. The magnitude of 
velocity in circular motion remains 
constant but the direction changes 
continuously. 

Let us consider a particle of mass 

m moving with a velocity v along the circle 

of radius r with centre O as shown in Fig 

2.36. P is the position of the particle at a given instant of time such 

that the radial line OP makes an angle 8 with the reference line DA. The 

magnitude of the velocity remains constant, but its direction changes 

continuously. The linear velocity always acts tangentially to the position 

of the particle (i.e) in each position, the linear velocity v is perpendicular 

— > 
to the radius vector r. 




Fig. 2.36 Uniform circular motion 




2.6.1 Angular displacement 

Let us consider a particle of mass m 
moving along the circular path of radius r as 
shown in Fig. 2.37. Let the initial position of 
the particle be A. P and Q are the positions of 
| A the particle at any instants of time t and t + dt 
respectively. Suppose the particle traverses a 
distance ds along the circular path in time 
interval dt. During this interval, it moves through 
an angle d9 = 9 2 - 9 1 . The angle swept by the 
radius vector at a given time is called the angular 
displacement of the particle. 

If r be the radius of the circle, then the angular displacement is 

ds 
given by d8 = — . The angular displacement is measured in terms of 

radian. 

2.6.2 Angular velocity 

The rate of change of angular displacement is called the angular 
velocity of the particle. 



Fig. 2.37 Angular 
displacement 



80 



Let dO be the angular displacement made by the particle in 

dO 



time dt , then the angular velocity of the particle is to = — . Its unit 

is rad s~ J and dimensional formula is T~ J . 

For one complete revolution, the angle swept by the radius vector 
is 360° or 2n radians. If Tis the time taken for one complete revolution, 

6 _2n 



known as period, then the angular velocity of the particle is to 



CO 



If the particle makes n revolutions per second, then 

— z,k n where n = — is the frequency of revolution. 
V J J 1 



2n 



2.6.3 Relation between linear velocity and angular velocity 

Let us consider a body P moving along the circumference of a 
circle of radius r with linear velocity v and angular velocity co as shown 
in Fig. 2.38. Let it move from P to Q in time dt and dO be the angle 
swept by the radius vector. 

Let PQ = ds, be the arc length covered 
by the particle moving along the circle, then 
the angular displacement d 8 is expressed 

ds 

— . But ds = v dt 
r 

v dt , , de v 

(nri = — 

dt r 



as d6 = 



dO = 



or 




i.e 



Angular velocity co = — or v =a> r 



In vector notation, v 



a x r 



Fig 2.38 Relation 

between linear velocity 

and angular velocity 



Thus, for a given angular velocity co, the linear velocity v of the 
particle is directly proportional to the distance of the particle from the 
centre of the circular path (i.e) for a body in a uniform circular motion, 
the angular velocity is the same for all points in the body but linear 
velocity is different for different points of the body. 

2.6.4 Angular acceleration 

If the angular velocity of the body performing rotatory motion is 
non-uniform, then the body is said to possess angular acceleration. 

81 



The rate of change of angular velocity is called angular acceleration. 

If the angular velocity of a body moving in a circular path changes 
from coj to co 2 in time t then its angular acceleration is 

dm dfd9\ d 2 6 m 2 -m 1 
"~ dt ~ dt{dt)~ dt 2 ~ t 

The angular acceleration is measured in terms of rad s~ 2 and its 
dimensional formula is T 2 . 

2.6.5 Relation between linear acceleration and angular 
acceleration 

If dv is the small change in linear velocity in a time interval dt 

dv d i n dm 
then linear acceleration is o. = — — = — (rm) = r —— = ra . 

dt dt dt 

2.6.6 Centripetal acceleration 

The speed of a particle performing uniform circular motion remains 
constant throughout the motion but its velocity changes continuously 
due to the change in direction (i.e) the particle executing uniform circular 
motion is said to possess an acceleration. 

Consider a particle executing circular motion of radius r with 
linear velocity v and angular velocity co. The linear velocity of the 
particle acts along the tangential line. Let dO be the angle described 
by the particle at the centre whe:i it moves from A to B in time dt. 

At A and B, linear velocity v acts along 
AH and BT respectively. In Fig. 2.39 
ZAOB =d6 = ZHET (v angle subtended by 
the two radii of a circle = angle subtended 
by the two tangents). 

The velocity u at B of the particle 
makes an angle d8 with the line BC and 
hence it is resolved horizontally as v cos d0 

along BC and vertically as v sin d 6 along Fig 2 .39 Centripetal 

BD. acceleration 

:. The change in velocity along the horizontal direction = v cos dd -v 

If dO is very small, cos dO = 1 

82 




Change in velocity along the horizontal direction = v - v = 
(i.e) there is no change in velocity in the horizontal direction. 
The change in velocity in the vertical direction (i.e along AO) is 
dv = v sin d6 - = v sin d6 
If dO is very small, sin dQ = dQ 
.'. The change in velocity in the vertical direction (i.e) along radius 



of the circle 

dv = v.dO 



where co 



But, acceleration a 

ae 



dv 
~dt 



vd6 
dt 



V(d 



dt 



is the angular velocity of the particle. 



We know that v = r co 
From equations (2) and (3), 

2 V 2 

a = rco co = rco z = — 



...d) 
...(2) 

...(3) 
• ••(4) 



Hence, the acceleration of the particle producing uniform circular 

v 2 
motion is equal to — and is along AO (i.e) directed towards the centre of 
r 

the circle. This acceleration is directed towards the centre of the circle 
along the radius and perpendicular to the velocity of the particle. This 
acceleration is known as centripetal or radial or normal acceleration. 

2.6.7 Centripetal force 

According to Newton's first law of motion, a body possesses the 
property called directional inertia (i.e) the inability of the body to change 
its direction. This means that without the 
application of an external force, the direction 
of motion can not be changed. Thus when 
a body is moving along a circular path, 
some force must be acting upon it, which 
continuously changes the body from its 
straight-line path (Fig 2.40). It makes clear 
that the applied force should have no 
component in the direction of the motion of v 
the body or the force must act at every Fig 2.40 Centripetal force 




83 



point perpendicular to the direction of motion of the body. This force, 
therefore, must act along the radius and should be directed towards the 
centre. 

Hence for circular motion, a constant force should act on the body, 
along the radius towards the centre and perpendicular to the velocity 
of the body. This force is known as centripetal force. 

If m is the mass of the body, then the magnitude of the centripetal 

force is given by 

F = mass x centripetal acceleration 

fv 2 '] mv 2 . 2 \ 

= m — = = m (rar) 

Examples 

Any force like gravitational force, frictional force, electric force, 
magnetic force etc. may act as a centripetal force. Some of the examples 
of centripetal force are : 

(i) In the case of a stone tied to the end of a string whirled in a 
circular path, the centripetal force is provided by the tension in the 
string. 

(ii) When a car takes a turn on the road, the frictional force 
between the tyres and the road provides the centripetal force. 

(iii) In the case of planets revolving round the Sun or the moon 
revolving round the earth, the centripetal force is provided by the 
gravitational force of attraction between them 

(iv) For an electron revolving round the nucleus in a circular 
path, the electrostatic force of attraction between the electron and the 
nucleus provides the necessary centripetal force. 

2.6.8 Centrifugal reaction 

According to Newton's third law of motion, for every action there 
is an equal and opposite reaction. The equal and opposite reaction to the 
centripetal force is called centrifugal reaction, because it tends to take the 
body away from the centre. In fact, the centrifugal reaction is a pseudo 
or apparent force, acts or assumed to act because of the acceleration 
of the rotating body. 

In the case of a stone tied to the end of the string is whirled in 
a circular path, not only the stone is acted upon by a force (centripetal 
force) along the string towards the centre, but the stone also exerts an 
equal and opposite force on the hand (centrifugal force) away from the 

84 



centre, along the string. On releasing the string, the tension disappears 
and the stone flies off tangentially to the circular path along a straight 
line as enuciated by Newton's first law of motion. 

When a car is turning round a corner, the person sitting inside 
the car experiences an outward force. It is because of the fact that no 
centripetal force is supplied by the person. Therefore, to avoid the 
outward force, the person should exert an inward force. 



2.6.9 Applications of centripetal forces 

(i) Motion in a vertical circle 

Let us consider a body of mass m 
tied to one end of the string which is 
fixed at O and it is moving in a vertical 
circle of radius r about the point O as 
shown in Fig. 2.41. The motion is circular 
but is not uniform, since the body speeds 
up while coming down and slows down 
while going up. 

Suppose the body is at P at any 
instant of time t, the tension T in the 
string always acts towards 0. 

The weight mg of the body at P is 
resolved along the string as mg cos 8 
which acts outwards and mg sin 8, 
perpendicular to the string. 

When the body is at P, the following 
forces acts on it along the string. 

(i) mg cos 8 acts along OP (outwards) 

(ii) tension T acts along PO (inwards) 

Net force on the body at P acting along PO = T - mg cos 8 




mg cos9 



Fig. 2.41 Motion of a body 
in a vertical circle 



This must provide the necessary centripetal force 



mv 



Therefore, T - mg cos 8 = 



T = mg cos 8 + 



• ••(I) 



85 



At the lowest point A of the path, 6 = 0°, cos 0° = 1 then 

mv 2 . 



from equation 


(1), T A 


= mg + — 




...(2) 


At the highest point of the path, i.e. at B, = 


180°. 


Hence 


cos 180°= -1 






.". from equation (1), T R - - mg + - 

r r 


- mg 




f" 2 ^ 






T B = m 


-^-9 

v r 






...(3) 



If T B > 0, then the string remains taut while if T B < 0, the string 
slackens and it becomes impossible to complete the motion in a vertical 
circle. 

If the velocity v B is decreased, the tension T B in the string also 
decreases, and becomes zero at a certain minimum value of the speed 
called critical velocity. Let v c be the minimum value of the velocity, 
then at v B = v c , T B = 0. Therefore from equation (3), 

.2 



mv r 



mg = (or) v 2 c = rg 



r 
(i.e) v c = jrg ...(4) 

If the velocity of the body at the highest point B is below this 
critical velocity, the string becomes slack and the body falls downwards 
instead of moving along the circular path. In order to ensure that the 
velocity v B at the top is not lesser than the critical velocity Jrg , the 
minimum velocity v A at the lowest point should be in such a way that 
v B should be Jrg . (i.e) the motion in a vertical circle is possible only 
if v B >jrg. 

The velocity v A of the body at the bottom point A can be obtained 
by using law of conservation of energy. When the stone rises from A 
to B, i.e through a height 2r, its potential energy increases by an 
amount equal to the decrease in kinetic energy. Thus, 

(Potential energy at A + Kinetic energy at A ) = 

(Potential energy at B + Kinetic energy at B) 

(i.e.) + — m v 2 A = mg (2r) +- m v B 

Dividing by j, v A = v B + 4gr ...(5) 

86 



But from equation (4), v\ = gr ('.' v B =V C ) 

.•. Equation (5) becomes, v A = gr + 4gr (or) v A = 

Substituting v A from equation (6) in (2), 
m[5gr) 



T.= mg + 



= mg + 5mg = 6 mg 



...(6) 



.(7) 



While rotating in a vertical circle, the stone must have a velocity 
greater than *j5gr or tension greater than 6mg at the lowest point, so 
that its velocity at the top is greater than Jgr or tension > 0. 

An aeroplane while looping a vertical circle must have a velocity 
greater than ^5gr at the lowest point, so that its velocity at the top is 
greater than Jgr. In that case, pilot sitting in the aeroplane will not fall. 



(ii) Motion on a level circular road 

When a vehicle goes round a level 
curved path, it should be acted upon by 
a centripetal force. While negotiating the 
curved path, the wheels of the car have 
a tendency to leave the curved path and 
regain the straight-line path. Frictional 
force between the tyres and the road 
opposes this tendency of the wheels. This 
frictional force, therefore, acts towards the 
centre of the circular path and provides 
the necessary centripetal force. 




mg 

Fig. 2.42 Vehicle on a 
level circular road 



In Fig. 2.42, weight of the vehicle mg acts vertically downwards. 
R 2 are the forces of normal reaction of the road on the wheels. As 



the road is level (horizontal), R,, R 
R 



v r 



2 act vertically upwards. Obviously, 
1 + R 2 = mg ...(1) 

Let u * be the coefficient of friction between the tyres and the 



*Friction : Whenever a body slides over another body, a force comes into play 
between the two surfaces in contact and this force is known as frictional force. 
The frictional force always acts in the opposite direction to that of the motion of 
the body. The frictional force depends on the normal reaction. (Normal reaction is 
a perpendicular reacUonal force that acts on the body at the point of contact due 
to its own weight) (i.e) Frictional force a normal reaction F a R (or) F = iiR where \x 
is a proportionality constant and is known as the coefficient of friction. The 
coefficient of friction depends on the nature of the surface. 



87 



road, F 1 and F 2 be the forces of friction between the tyres and the road, 
directed towards the centre of the curved path. 

.-. Fj = / uR 1 and F 2 = yR 2 ...(2) 

If v is velocity of the vehicle while negotiating the curve, the 

9 
mir 

centripetal force required = 



r 
As this force is provided only by the force of friction. 

, ^(F 1+ F 2 ) 
r 
<(HR 1 + nR 2 ) 

< n (Rj + R 2 ) 

<nmg (■; R 1 +R 2 =mg) 



mi) 2 



v 2 < n rg 



v<4mt9 

Hence the maximum velocity with which a car can go round a 
level curve without skidding is u = ^jjtrg . The value of v depends on 
radius r of the curve and coefficient of friction |i between the tyres and 
the road. 

(Hi) Banking of curved roads and tracks 

When a car goes round a level curve, the force of friction between 
the tyres and the road provides the necessary centripetal force. If the 
frictional force, which acts as centripetal force and keeps the body 
moving along the circular road is not enough to provide the necessary 
centripetal force, the car will skid. In order to avoid skidding, while 
going round a curved path the outer edge of the road is raised above 
the level of the inner edge. This is known as banking of curved roads 
or tracks. 

Bending of a cyclist round a curve 

A cyclist has to bend slightly towards the centre of the circular 
track in order to take a safe turn without slipping. 

Fig. 2.43 shows a cyclist taking a turn towards his right on a 
circular path of radius r. Let m be the mass of the cyclist along with 
the bicycle and u, the velocity. When the cyclist negotiates the curve, 
he bends inwards from the vertical, by an angle 0. Let R be the reaction 

88 




Rcos e 



i 

mg 



■> F 




R sin 6 



Fig 2.43 Bending of a cyclist in a curved road 

of the ground on the cyclist. The reaction R may be resolved into two 
components: (i) the component R sin 0, acting towards the centre of the 
curve providing necessary centripetal force for circular motion and 
(ii) the component R cos 9, balancing the weight of the cyclist along 
with the bicycle. 



(i.e) 
and 



R sin 
R cos 8 -- 



mv 



mg 



Dividing equation (1) by (2), 



R sin 6 
R cos e 



mv 

r 

mg 



...(1) 
.-(2) 



tan 6 = 



rg 



...(3) 



Thus for less bending of cyclist (i.e for to 
be small), the velocity v should be smaller and 
radius r should be larger. 

For a banked road (Fig. 2.44), let h be the 
elevation of the outer edge of the road above the 

inner edge and I be the width of the road then, 

h 

T 



sin H = 



...(4) 




Fig 2.44 Banked road 



89 



For small values of 8, sin 8 = tan 8 
Therefore from equations (3) and (4) 



h v 



• 



tan 6 = J = — ...(5) 

Obviously, a road or track can be banked correctly only for a 
particular speed of the vehicle. Therefore, the driver must drive with 
a particular speed at the circular turn. If the speed is higher than the 
desired value, the vehicle tends to slip outward at the turn but then 
the frictional force acts inwards and provides the additional centripetal 
force. Similarly, if the speed of the vehicle is lower than the desired 
speed it tends to slip inward at the turn but now the frictional force 
acts outwards and reduces the centripetal force. 

Condition for skidding 

When the centripetal force is greater than the frictional force, 
skidding occurs. If \i is the coefficient of friction between the road and 
tyre, then the limiting friction (frictional force) is f = /j.R where normal 
reaction R = mg 

:.f = n (mg) 

Thus for skidding, 

Centripetal force > Frictional force 

m ii 

> H (mg) 



r 

,2 



> J" 



V 

But — = tan 

rg 

:. tan 6 > /j. 

(i.e) when the tangent of the angle of banking is greater than the 
coefficient of friction, skidding occurs. 

2.7 Work 

The terms work and energy are quite familiar to us and we use 
them in various contexts. In everyday life, the term work is used to 
refer to any form of activity that requires the exertion of mental or 
muscular efforts. In physics, work is said to be done by a force or 

90 



against the direction of the force, when the point of application of the 
force moves towards or against the direction of the force. If no 
displacement takes place, no work is said to be done. Therefore for 
work to be done, two essential conditions should be satisfied: 

(i) a force must be exerted 

(ii) the force must cause a motion or displacement 

If a particle is subjected to a force F and if the particle is displaced 
by an infinitesimal displacement ds , the work done dw by the force is 
dw = 

The magnitude of the above dot product 
is Fcos 8 ds. 

(le) dw = Fds cos 8 = (Fcos 0) ds where 
8 = angle between F and as. (Fig. 2.45) 




>Pi 



Thus, the work done by a force during an 

infinitesimal displacement is equal to the product 

of the displacement ds and the component of 
Fig. 2.45 Work done _,, , _ n . _,, ,. , _,, 

, r- the force F cos 6 in the direction of the 

by a Jorce J J 

displacement. 

Work is a scalar quantity and has magnitude but no direction. 

The work done by a force when the body is displaced from position 
P to P 1 can be obtained by integrating the above equation, 

W= I dw = j (F cos 8 ) ds 

Work done by a constant force 

When the force F acting on a body 

has a constant magnitude and acts at a 

constant angle 8 from the straight line o 

path of the particle as shown as Fig. 

2.46, then, 

s 2 
„ „ f , „ , , Fiq. 2.46 Work done by a 

W=Fcos9\ ds = FcosB(s 2 - Sl ) constant force 

s , 
The graphical representation of work done by a constant force is 

shown in Fig 2.47. 

W = F cos 6 (s 2 -Sj) = area ABCD 

91 




Y 




-»x 



O s, s 2 s 

Fig. 2. 47 Graphical representation 
of work done by a constant force 




O s i a b s 2 s 

Fig 2.48 Work done by a 

variable force 



Work done by a variable force 

If the body is subjected to a varying force F and displaced along 
X axis as shown in Fig 2.48, work done 

dw = Fcos 8. ds = area of the small element abed. 

.*. The total work done when the body moves from Sj to s 2 is 

E dw= W = area under the curve P 1 P 2 = area S_, P 1 P 2 S 2 

The unit of work is joule. One joule is defined as the work done 

by a force of one newton when its point of application moves by one 

metre along the line of action of the force. 

Special cases 

(i) When 8 = 0, the force F is in the same direction as the 
displacement s. 

.■. Work done, W = F s cos = F s 

(ii) When 8 = 90°, the force under consideration is normal to the 
direction of motion. 

/.Work done, W = F s cos 90° = 

For example, if a body moves along africtionless horizontal surface, 
its weight and the reaction of the surface, both normal to the surface, do 
no work. Similarly, when a stone tied to a string is whirled around in a 
circle with uniform speed, the centripetal force continuously changes the 
direction of motion. Since this force is always normal to the direction of 
motion of the object, it does no work. 

(iii) When 8 = 180°, the force Fis in the opposite direction to the 
displacement. 

92 



.-. Work done (W)=Fs cos 180°= -F s 

(eg.) The frictional force that slows the sliding of an object over 
a surface does a negative work. 

A positive work can be defined as the work done by a force and 
a negative work as the work done against a force. 

2.8 Energy 

Energy can be defined as the capacity to do work. Energy can 
manifest itself in many forms like mechanical energy, thermal energy, 
electric energy, chemical energy, light energy, nuclear energy, etc. 

The energy possessed by a body due to its position or due to its 
motion is called mechanical energy. 

The mechanical energy of a body consists of potential energy and 
kinetic energy. 

2.8.1 Potential energy 

The potential energy of a body is the energy stored in the body by 
virtue of its position or the state of strain. Hence water stored in a 
reservoir, a wound spring, compressed air, stretched rubber chord, etc, 
possess potential energy. 

Potential energy is given by the amount of work done by the force 
acting on the body, when the body moves from its given position to 
some other position. 

Expression for the potential energy 

Let us consider a body of mass m, which is at 
rest at a height h above the ground as shown in mg 

Fig 2.49. The work done in raising the body from the 
ground to the height h is stored in the body as its 
potential energy and when the body falls to the ground, MUMUMMMMfjM 
the same amount of work can be got back from it. Fig. 2.49 

Now, in order to lift the body vertically up, a force mg Potential energy 
equal to the weight of the body should be applied. 

When the body is taken vertically up through a height h, then 
work done, W = Force x displacement 
.". W = mg x h 

This work done is stored as potential energy in the body 
/. E p = mgh 

93 



h 



2.8.2 Kinetic energy 

The kinetic energy of a body is the energy possessed by the body 
by virtue of its motion. It is measured by the amount of work that the 
body can perform against the impressed forces before it comes to rest. 
A falling body, a bullet fired from a rifle, a swinging pendulum, etc. 
possess kinetic energy. 

A body is capable of doing work if it moves, but in the process 
of doing work its velocity gradually decreases. The amount of work that 
can be done depends both on the magnitude of the velocity and the 
mass of the body. A heavy bullet will penetrate a wooden plank deeper 
than a light bullet of equal size moving with equal velocity. 

Expression for Kinetic energy 

Let us consider a body of mass m moving with a velocity v in a 
straightline as shown in Fig. 2.50. Suppose that it is acted upon by a 
constant force F resisting its motion, which produces retardation a 
(decrease in acceleration is known as retardation). Then 

F = mass x retardation = - ma ...(1) 

Let dx be the displacement of the 
2 ^i body before it comes to rest. 



v 



I But the retardation is 

I — I — 1 
I • I 



F * dv dv dx dv 

a = — = — x — = — xv ...(21 
1 1- dt dx dt dx l ' 



( rest > dx 

Fig. 2.50 Kinetic energy where & = V iS the Vel ° Clty ° f the b ° dy 

Substituting equation (2) in (1), F = - mv — ...(3) 

Hence the work done in bringing the body to rest is given by, 

o dv , o 

W= jF.dx = -\mv. — .dx = - m \ v dv ...(4) 



dx 

2 1° 



W = -m 



This work done is equal to kinetic energy of the body. 



94 



Kinetic energy E fc 



mrr 



2.8.3 Principle of work and energy (work - energy theorem) 

Statement 

The work done by a force acting on the body during its displacement 
is equal to the change in the kinetic energy of the body during that 
displacement. 

Proof 

Let us consider a body of mass m acted upon by a force F and 
moving with a velocity v along a path as shown in Fig. 2.51. At any 
instant, let P be the position of the body 
from the origin O. Let 8 be the angle made 
by the direction of the force with the 
tangential line drawn at P. 

The force F can be resolved into two 
rectangular components : 

(i) F t = F cos 8 , tangentially and 

(ii) F ' = F sin 8 , normally at P. 



But F t = ma t 




.(1) 



where a t is the acceleration of the body in 
the tangential direction 

F cos 8 



ma. 



But a. 



dv 
~dt 



Fig. 2.51 
Work-energy theorem 



.-(2) 
.-(3) 



.". substituting equation (3) in (2), 
F cos 8 



dv dv ds 

ds dt 



.-(4) 
...(5) 



dt 

F cos8 ds = mv dv 
where ds is the small displacement. 

Let Uj and v 2 be the velocities of the body at the positions 1 and 2 
and the corresponding distances be s 1 and s 2 . 

Integrating the equation (5), 



J (F cos 8) ds = J mv dv 



...(6) 



95 



S 2 

But \ [Fcos 9) ds = W^ 2 ...(7) 

where W } 2 is the work done by the force 
From equation (6) and (7), 

V 2 

W 1 _ >2 = I mv dv 



v 2 



mv 2 mv 2 



• (8) 



-<v. 



= m 

Therefore work done 

= final kinetic energy - initial kinetic energy 
= change in kinetic energy 
This is known as Work-energy theorem. 

2.8.4 Conservative forces and non-conservative forces 

Conservative forces 

If the work done by a force in moving a body between two positions 
is independent of the path followed by the body, then such a force is 
called as a conservative force. 

Examples : force due to gravity, spring force and elastic force. 

The work done by the conservative forces depends only 
upon the initial and final position of the body. 

(i.e.) j F . dr = 

The work done by a conservative force around a closed path is 
zero. 

Non conservative forces 

Non-conservative force is the force, which can perform some 
resultant work along an arbitrary closed path of its point of application. 

The work done by the non-conservative force depends upon the 
path of the displacement of the body 



96 



(i.e.) j> F . dr * 

(e.g) Frictional force, viscous force, etc. 

2.8.5 Law of conservation of energy 

The law states that, if a body or system of bodies is in motion under a 
conservative system of forces, the sum of its kinetic energy and potential 
energy is constant. 

Explanation 

From the principle of work and energy, 

Work done = change in the kinetic energy 

(i.e)W^ 2 = E k2 -E kl ...(1) 

If a body moves under the action of a conservative force, work 
done is stored as potential energy. 

W 1^2 = - ( E P2 ~ E Pl) -(2) 

Work done is equal to negative change of potential energy. 
Combining the equation (1) and (2), 

Ek 2 - Ek 1 = -(E p2 - E pi ) (or) E pi + E kl = E p2 + E k2 ...(3) 

which means that the sum of the potential energy and kinetic energy of 
a system of particles remains constant during the motion under the action 
of the conservative forces. 

2.8.6 Power 

It is defined as the rate at which work is done. 

work done 



power = 

time 

Its unit is watt and dimensional formula is ML 2 T~ 3 . 

Power is said to be one watt, when one joule of work is said to be 
done in one second. 

If dw is the work done during an interval of time dt then, 

dw 
power = -^ ...(1) 

But dw = (F cos 9) ds ...(2) 

97 



where 8 is the angle between the direction of the force and displacement. 
F cos is component of the force in the direction of the small 
displacement ds. 

(F cos 9) ds 
Substituting equation (2) in (1) power = — 

= (Fcos9) — = (F cos 9) v [•: — = 

dt { dt 

:. power = (F cos 9) u 

If F and v are in the same direction, then 
power = F v cos = F v = Force x velocity 

It is also represented by the dot product of F and v. 

— > -> 
(i.e) P = F . v 

2.9 Collisions 

A collision between two particles is said to occur if they physically 
strike against each other or if the path of the motion of one is influenced 
by the other. In physics, the term collision does not necessarily mean 
that a particle actually strikes. In fact, two particles may not even 
touch each other and yet they are said to collide if one particle influences 
the motion of the other. 

When two bodies collide, each body exerts a force on the other. 
The two forces are exerted simultaneously for an equal but short interval 
of time. According to Newton's third law of motion, each body exerts an 
equal and opposite force on the other at each instant of collision. 
During a collision, the two fundamental conservation laws namely, the 
law of conservation of momentum and that of energy are obeyed and 
these laws can be used to determine the velocities of the bodies after 
collision. 

Collisions are divided into two types : (i) elastic collision and 
(ii) inelastic collision 

2.9.1 Elastic collision 

If the kinetic energy of the system is conserved during a collision, 
it is called an elastic collision, (i.e) The total kinetic energy before collision 
and after collision remains unchanged. The collision between subatomic 

98 



particles is generally elastic. The collision between two steel or glass 
balls is nearly elastic. In elastic collision, the linear momentum and 
kinetic energy of the system are conserved. 

Elastic collision in one dimension 

If the two bodies after collision move in a straight line, the collision 
is said to be of one dimension. 

Consider two bodies A and B of masses m 1 and m 2 moving along 
the same straight line in the same direction with velocities u 1 and u 2 
respectively as shown in Fig. 2.54. Let us assume that u 1 is greater than 

u 2 . The bodies A and B suffer 
a head on collision when 
they strike and continue to 
move along the same straight 
line with velocities v l and v 2 




(Twa^ 



respectively. 






\^>/ From the law of 

conservation of linear 
Fig 2.54 Elastic collision in one dimension momentum 

Total momentum before collision = 

Total momentum after collision 
m 1 u 1 + m 2 u 2 = m 1 v 1 + m 2 v 2 ...(1) 

Since the kinetic energy of the bodies is also conserved during the 
collision 

Total kinetic energy before collision = 

Total kinetic energy after collision 

- m i"i 2 + ~ m 2 u l = - m i v l + ~ m 2 v l -..(2) 

rrijUj 2 - rrijuf = m 2 v 2 - m 2 u% ...(3) 

From equation (1) m l (u l - v 1 ) = m 2 (u 2 - u 2 ) ■■■{4) 

Dividing equation (3) by (4), 

(or) Uj + Uj = u 2 + v 2 



u?-v? 



(u 1 - u 2 ) = (v 2 - Vj) ...(5) 

99 



Equation (5) shows that in an elastic one-dimensional collision, 
the relative velocity with which the two bodies approach each other 
before collision is equal to the relative velocity with which they recede 
from each other after collision. 

From equation (5), v 2 = u 1 - u 2 + v 1 ...(6) 

Substituting v 2 in equation (4), 
rrij ( Uj- Vj) = m 2 ( v 1 - u 2 + u 1 - u 2 ) 
m 1 u 1 - rrijU, = m 2 u 1 - 2m 2 u 2 + m 2 v 1 
(m 1 + m 2 )v 1 = m 1 u 1 - m 2 u 1 + 2m 2 u 2 
(m 1 + m 2 )v 1 = u 1 (m 1 - m 2 ) + 2m 2 u 2 

+ f7) 

rr\ +DT2 J {rr^+nr^) '" l ' 

2m Y Uj u 2 (m 2 - mj 
Similarly, v 2 = {mi + ^ + (rTli+m2 ) - (8) 

Special cases 

Case ( i) : If the masses of colliding bodies are equal, i.e. m 1 = m 2 
Uj = u 2 and v 2 = u 1 ...(9) 

After head on elastic collision, the velocities of the colliding bodies 
are mutually interchanged. 

Case (ii) : If the particle B is initially at rest, (i.e) u 2 = then 

(m A - m B ) 

and v 2 = ( 2mA u 1 ...(11) 

2.9.2 Inelastic collision 

During a collision between two bodies if there is a loss of kinetic 
energy, then the collision is said to be an inelastic collision. Since there 
is always some loss of kinetic energy in any collision, collisions are 
generally inelastic. In inelastic collision, the linear momentum is 
conserved but the energy is not conserved. If two bodies stick together, 
after colliding, the collision is perfectly inelastic but it is a special case 
of inelastic collision called plastic collision, (eg) a bullet striking a block 

100 



of wood and being embedded in it. The loss of kinetic energy usually 
results in the form of heat or sound energy. 

Let us consider a simple situation in which the inelastic head on 
collision between two bodies of masses m A and m B takes place. Let the 
colliding bodies be initially move with velocities u 1 and u 2 . After collision 
both bodies stick together and moves with common velocity v. 

Total momentum of the system before collision = m A u 1 + m B u 2 

Total momentum of the system after collision = 

mass of the composite body x common velocity = (m A + m B ) v 
By law of conservation of momentum 

m A u A +m B u B 



m A u 1 + m B u 2 = (m A + m B ) v (or) u = 



m A +m B 



Thus, knowing the masses of the two bodies and their velocities 
before collision, the common velocity of the system after collision can 
be calculated. 

If the second particle is initially at rest i.e. u 2 = then 



(m A +m B ) 
kinetic energy of the system before collision 



1 2 

—mA 1 

2 
and kinetic energy of the system after collision 



E K1 = n m ^ ' ••• U 2 



Hence, 



E K2 = oK + ^l" 

kinetic energy after collision 
kinetic energy before collision 



(m A + m B )u 



2 



" L A"-A 

Substituting the value of v in the above equation, 

E K2 _ m A E K2 

E Kl m A +m B lor) E K1 < 

It is clear from the above equation that in a perfectly inelastic 
collision, the kinetic energy after impact is less than the kinetic energy 
before impact. The loss in kinetic energy may appear as heat energy. 

101 



Solved Problems 

2. 1 The driver of a car travelling at 72 kmph observes the light 

300 m ahead of him turning red. The traffic light is timed to 
remain red for 20 s before it turns green. If the motorist 
wishes to passes the light without stopping to wait for it to 
turn green, determine (i) the required uniform acceleration of 
the car (ii) the speed with which the motorist crosses the 
traffic light. 

Data : u = 72 kmph = 72 x -^ m s " 1 = 20 m s ~* ; S= 300 m ; 



t= 20s ; a=? 

c = lit 4- 

2 



Solution : i) s = ut + — at 2 



300 =(20x20) + - a (20) 2 

a = - 0.5 ms~ 2 

ii) v = u + at = 20 - 0.5 x 20 = 10 m s" J 

2.2 A stone is dropped from the top of the tower 50 m high. At the 

same time another stone is thrown up from the foot of the 
tower with a velocity of 25 m s _ 1 . At what distance from the 
top and after how much time the stones cross each other? 

Data: Height of the tower = 50 m u 1 = ; u 2 = 25 m s~ 1 

Let s 1 and s 2 be the distances travelled by the two stones at the 
time of crossing (t). Therefore 5^52 = 50m 



Sl =?;t=? 




Solution : For I stone : 


s^^gt 2 


For II stone : 


1 .9 
S 2 = U 2 t -2 g t 




s 2 = 25t-\gt 2 


Therefore, S}+s 2 


= 50=\gt 2 +25t-\gt 2 



t= 2 seconds 

s 1 = \gt? = \ (9.8) (2) 2 = 19.6 m 



102 



2.3 A boy throws a ball so that it may just clear a wall 3.6m high. 

The boy is at a distance of 4.8 m from the wall. The ball was 
found to hit the ground at a distance of 3.6m on the other side 
of the wall. Find the least velocity with which the ball can be 
thrown. 

Data: Range of the ball = 4.8 + 3.6 =8.4m 

Height of the wall = 3.6m 

u = ?; e=? 

Solution : The top of the wall AC must lie on the path of the 
projectile. 



The equation of the projectile is y = x tan 6 



9* 



2u 2 cos 2 e 



The point C (x = 4.8m, u = 3.6m) lies on the trajectory. 
Substituting the known values in (1), 

,2 



3.6 =4.8 tarn 



g x (4.8V 

9 O 

2u COS 



.(l) 



...(2) 



The range of the projectile is R 



u^ sin 29 
9 



8.4 



...(3) 



o 

<- 



From (3), 



4.8 m 
.2 



u 



8.4 



g sin 29 
Substituting (4) in (2), 



> 




C 

\ ' B 
\ i ^° 


U. 




\ i ^ 




i 


V 



- J ^- 



3.6 m 



...(4) 



3.6 =(4.8) tan 9 



(4.8V 



sin 29 



2 cos 2 9 (8.4) 



103 



Kir ,ab\+ a < 4 - 8 ) 2 2sin0cos0 
3.6 = (4.8) tan — x 

2 cos 2 9 (8A) 

3.6 = (4.8) tan - (2.7429) tan 6 



Substituting the value of in (4 ), 
2 8.4 xg 8.4x9.8 



sin 20 sin 2(60° 15') 
u =9.7745 ms 1 



95.5399 



2.4 Prove that for a given velocity of projection, the horizontal 

range is same for two angles of projection a and (90° - a}. 

_ u 2 sin 20 
The horizontal range is given by, R = ...(1) 

When 8 = a, 

u 2 sin 2a gg J ggjgjgfr^ fl u 2 sin 2a 
R l~ - *2" 2M57J " - -(2) 

When = (90° -a), = tan -1 [i .75] = 60°1 5' 

_ u 2 sin 2(90°- a) _ u 2 [2 sin(90°- a) cos(90°- a] 
2 9 9 '" 

But sin(90 - a) = cos a ; cos(90 - a) = sin a 



...(4) 

From (2) and (4), it is seen that at both angles a and (90 -a), the 
horizontal range remains the same. 

2.5 The pilot of an aeroplane flying horizontally at a height of 

2000 m with a constant speed of 540 kmph wishes to hit a 
target on the ground. At what distance from the target should 
release the bomb to hit the target? 



104 



Data : Initial velocity of the bomb in the horizontal is the same 
as that of the air plane. 

Initial velocity of the bomb in the horizontal 



direction = 540 kmph = 540 x — - m s - 

lo 



150 ms- 1 



Initial velocity in the vertical direction (u) = ; vertical distance 
(s) = 2000 m ; time of flight t = ? 

Solution : From equation of motion, 

1 9 

s = ut + —at 
2 

Substituting the known values, 



2000 =0xt + -x9.8xt 2 
2 



2000=4.9t' 



(or) 




t = 



2000 
4.9 



= 20.20 s 



R Target B 

> 



.". horizontal range = horizontal velocity x time of flight 
= 150 x 20.20 = 3030 m 

2.6 Two equal forces are acting at a point with an angle of 60° 

between them. If the resultant force is equal to 2(W3 N, find 
the magnitude of each force. 

Data : Angle between the forces, 6 = 60° ; Resultant R = 20^3 N 

P = Q = P (say) = ? 



Solution : R 



= ^P 2 +Q 2 +2PQ cos9 
= ^JP 2 +P 2 +2P.P cos 60° 



2P 2 +2P 2 . - = p ^ 



20 73 = P V3 
P =20N 

105 



2.7 



2.8 



If two forces F, = 20 fciV and F 2 = 15 kJV act on a particle as 
shown in figure, find their resultant by triangle law. 

Data : F 1 =20kN;F 2 = 15 k N; R=? 

Solution : Using law of cosines, 

R 2 = P 2 + Q 2 - 2PQ cos (180 - 0) 

R 2 = 20 2 + 15 2 - 2 (20) (15) cos 110° 

.: R = 28.813 kN. 

Using law of sines, ^,'' 

R 15 ,.'"' 

,'\ a 

^1 i_ 



sin 110 sin a 
a = 29.3° 




20 



Two forces act at a point in directions inclined to each other at 
120°. If the bigger force is 5 kg wt and their resultant is at 
right angles to the smaller force, find the resultant and the 
smaller force. 



Data : Bigger force = 5 kg wt 

Angle made by the resultant with the smaller force 

Resultant = ? Smaller force = ? 



90° 



Solution : Let the forces P and Q are acting along OA and OD 
where ZAOD =120° 

Complete the parallelogram OACD and join OC. OC therefore 
which represents the resultant which is perpendicular to OA. 

In AOAC 

ZOCA = ZCOD=30° 

ZAOC = 90° 
Therefore ZOAC = 60° 

He) P ~ Q ~ R 
sin 30 sin 90 sin 60 

Since Q = 5 kg. wt. 

5 sin 30 




R 



sin 90 

5 sin 60° 
sin 90° 



5 V3 



2.5 kg wt 



kg wt 



106 



2.9 Determine analytically the magnitude and direction of the 

resultant of the following four forces acting at a point. 

(i) 10 kN pull N 30° E; (ii) 20 kN push S 45° W; 

(iii) 5 kN push N 60° W; (iv) 15 kN push S 60° E. 

Data : F ; = 10 kN ; 



F 2 = 20kN ; 
F 3 = 5kN ; 
F 4 = 15 kN ; 



10 kN 



R 



9 ■ 




Solution : The various forces 
acting at a point are shown in 
figure. 

Resolving the forces 

horizontally, we get 

ZF x = 10 sin 30° + 5 sin 60° + 20 sin 45 

= 10.48 kN 

Similarly, resolving forces vertically, we get 

ZF y = 10 cos 30° - 5 cos 60° + 20 cos 45° + 15 cos 60° 

= 27.8 kN 

Resultant R = ^ (ZFJ 2 + (ZF y ) 2 



15 kN 



7&0.48) 2 + (27. 8) 2 
29.7 kN 



ZF„ 27.8 



tan a 



T.F x 10.48 
a = 69.34° 



2.65 



2.10 A machine weighing 1500 N is supported by two chains attached 
to some point on the machine. One of these ropes goes to a 
nail in the wall and is inclined at 30° to the horizontal and 



107 



other goes to the hook in ceiling 
and is inclined at 45° to the 
horizontal. Find the tensions in 
the two chains. 



Tensions 



Celling 

im 




Data: W=1500N. 
in the strings = ? 

Solution : The machine is in 
equilibrium under the following 
forces : 

(i) W ( weight of the machine) acting vertically down 

(ii) Tension T, in the chain OA; 

(Hi) Tension T 2 in the chain OB. 

Now applying Lami's theorem at O, we get 
T\ _ T 2 T 3 



V W= 1500 N 



sin (90° +45°} sin (90° +30°) sin 105° 



1500 



sin 135° sin 120° sin 105° 



T 1 500 x sin 135° = j 098 _ 96 N 

sin 105° 

T 1500xsin 120° = l34611 N 
sin 105° 

2.11 The radius of curvature of a railway line at a place when a 
train is moving with a speed of 72 kmph is 1500 m. If the 
distance between the rails is 1.54 m, find the elevation of the 
outer rail above the inner rail so that there is no side pressure 
on the rails. 

Data : r = 1500 m; v = 72 kmph= 20 ms' 1 ; I = 1.54 m; h=? 
h v 2 



Solution : 


tanO = — = — 
I rg 




Therefore 


lv 2 _ 1.54 x (20f 
h ~ rg 1500 x 9.8 


= 0.0419 m 



108 



2.12 A truck of weight 2 tonnes is slipped from a train travelling at 
9 kmph and comes to rest in 2 minutes. Find the retarding 
force on the truck. 

Data : m= 2 tonne = 2x1 000 kg = 2000 kg 

5 5 
v 1 = 9 kmph = 9 x — =— m s . ; v 2 = 

Solution : Let R newton be the retarding force. 
By the momentum - impulse theorem , 
( mv 1 - mv 2 ) = Rt (or) mu 1 - Rt = mv 2 

2000 x | -Rx 120 = 2000 x (or) 5000 - 120 R = 
R = 41.67 N 

2.13 A body of mass 2 kg initially at rest is moved by a horizontal 
force of 0.5N on a smooth frictionless table. Obtain the work 
done by the force in 8 s and show that this is equal to change 
in kinetic energy of the body. 

Data: M = 2 kg ; F = 0.5 N ; t=8s; W=? 

F 0.5 
Solution : .: Acceleration produced (a) = — = — — = 0.25 m s~ 2 

m 2 

The velocity of the body after 8s = ax t = 0.25 x 8 = 2 m s rl 

The distance covered by the body in8s = S = ut+-^ at 2 

S=(0x8) + - (0.25) (8) 2 = 8m 
2 

.'. Work done by the force in 8 s = 

Force x distance = 0.5 x 8 = 4J 

Initial kinetic energy = — m(0) 2 = 

1 o 1 
Final kinetic energy = — mv z = — x 2 x (2) z = 4 J 

^ 2 

.'. Change in kinetic energy = Final K.E. - Initial K.E = 4-0 = 4J 

The work done is equal to the change in kinetic energy of the 
body. 



109 



2.14 A body is thrown vertically up from the ground with a velocity 
of 39.2 m s _1 . At what height will its kinetic energy be reduced 
to one - fourth of its original kinetic energy. 

Data : v = 39.2 ms- 1 ;h=? 

Solution : When the body is thrown up, its velocity decreases 
and hence potential energy increases. 

Let h be the height at which the potential energy is reduced to 
one -fourth of its initial value. 

(i.e) loss in kinetic energy = gain in potential energy 

3 l 2 

x — mv z = mg h 



4 
3 



2 

1 



x - (39. 2) 2 = 9.8xh 



h= 58.8 m 



2.15 A 10 g bullet is fired from a rifle horizontally into a 5 kg block 
of wood suspended by a string and the bullet gets embedded in 
the block. The impact causes the block to swing to a height of 
5 cm above its initial level. Calculate the initial velocity of the 
bullet. 

Data : Mass of the bullet = m A = 10 g = 0.01 kg 
Mass of the wooden block = m B = 5 kg 
Initial velocity of the bullet before impact = u A = ? 
Initial velocity of the block before impact = u B = 
Final velocity of the bullet and block = v 



ss/ss/ss/s 




Bullet 



110 



Solution : By law of conservation of linear momentum, 

m A U A + m B U B = ( m A + "V U 

(0.01)u A + (5x0) = (0.01 + 5)v 

(0.01'] u A 

(or) U = l5^lJ U A = ^I -W 

Applying the law of conservation of mechanical energy, 

KE of the combined mass = PE at the highest point 

(or) - (m A + m B ) v 2 = (m A + m^ gh ...(2) 

From equation (1) and (2), 
ul 



(50 If 



2gh (or) u A = ^2.46 xlO 5 = 496.0 m s~ 



111 



Self evaluation 

(The questions and problems given in this self evaluation are only samples. 
In the same way any question and problem could be framed from the text 
matter. Students must be prepared to answer any question and problem 
from the text matter, not only from the self evaluation.) 

2.1 A particle at rest starts moving in a horizontal straight line with 
uniform acceleration. The ratio of the distance covered during 
the fourth and the third second is 

4 26 

(a) - (b) - 

(c) I (d) 2 

2.2 The distance travelled by a body, falling freely from rest in one, 
two and three seconds are in the ratio 

(a) 1:2:3 (b) 1:3:5 

(c) 1:4:9 (d) 9:4:1 

2.3 The displacement of the particle along a straight line at time t 
is given by, x = a Q + a 1 t +a 2 t 2 where a„,a 1 and a 2 are 
constants. The acceleration of the particle is 

(a) a (b) a 1 

(c) a 2 (d) 2a 2 

2.4 The acceleration of a moving body can be found from: 

(a) area under velocity-time graph 

(b) area under distance-time graph 

(c) slope of the velocity-time graph 

(d) slope of the distance-time graph 

2.5 Which of the following is a vector quantity? 

(a) Distance (b) Temperature 

(c) Mass (d) Momentum 

2.6 An object is thrown along a direction inclined at an angle 45° 
with the horizontal. The horizontal range of the object is 

(a) vertical height (b) twice the vertical height 

(c) thrice the vertical height (d) four times the vertical height 

112 



2.7 . Two bullets are fired at angle 8 and (90 - 8) to the horizontal with 

some speed. The ratio of their times of flight is 

(a) 1:1 (b) tan 8:1 

(c)l:tand (d) tan 2 8:1 

2.8 A stone is dropped from the window of a train moving along a 
horizontal straight track, the path of the stone as observed by an 
observer on ground is 

(a) Straight line (b) Parabola 

(c) Circular (c) Hyperbola 

2.9 A gun fires two bullets with same velocity at 60° and 30° with 
horizontal. The bullets strike at the same horizontal distance. 
The ratio of maximum height for the two bullets is in the ratio 

(a) 2 : 1 (b) 3 : 1 

(c) 4 : 1 (d) 1 : 1 

2.10 Newton's first law of motion gives the concept of 
(a) energy (b) work 

(c) momentum (d) Inertia 

2.11 Inertia of a body has direct dependence on 
(a) Velocity (b) Mass 

(c) Area (d) Volume 

2.12 The working of a rocket is based on 

(a) Newton' s first law of motion 

(b) Newton's second law of motion 

(c) Newton's third law of motion 

(d) Newton's first and second law 

2.13 When three forces acting at a point are in equilibrium 

(a) each force is equal to the vector sum of the other two forces. 

(b) each force is greater than the sum of the other two forces. 

(c) each force is greater than the difference of the other two 
force. 



113 



(d) each force is to product of the other two forces. 

2.14 For a particle revolving in a circular path, the acceleration of the 
particle is 

(a) along the tangent 

(b) along the radius 

(c) along the circumference of the circle 

(d) Zero 

2.15 If a particle travels in a circle, covering equal angles in equal 
times, its velocity vector 

(a) changes in magnitude only 

(b) remains constant 

(c) changes in direction only 

(d) changes both in magnitude and direction 

2.16 A particle moves along a circular path under the action of a 
force. The work done by the force is 

(a) positive and nonzero (b) Zero 

(c) Negative and nonzero (d) None of the above 

2.17 A cyclist of mass m is taking a circular turn of radius R on a 
frictional level road with a velocity v. Inorder that the cyclist 
does not skid, 

(a) (mv 2 /2) > nmg (b) (mv 2 /r) > ^img 

(c) (mrf/r) < nmg (d) (v/r) = /.ig 

2.18 If a force F is applied on a body and the body moves with 
velocity v, the power will be 

(a) F.v (b) F/v 

(c) Fv 2 (d) F/v 2 

2.19 For an elastic collision 

(a) the kinetic energy first increases and then decreases 

(b) final kinetic energy never remains constant 

(c) final kinetic energy is less than the initial kinetic energy 

114 



(d) initial kinetic energy is equal to the final kinetic energy 

2.20 A bullet hits and gets embedded in a solid block resting on a 
horizontal frictionless table. Which of the following is conserved? 

(a) momentum and kinetic energy 

(b) Kinetic energy alone 

(c) Momentum alone 

(d) Potential energy alone 

2.21 Compute the (i) distance travelled and (ii) displacement made by 
the student when he travels a distance of 4km eastwards and 
then a further distance of 3 km northwards. 

2.22 What is the (i) distance travelled and (ii) displacement produced 
by a cyclist when he completes one revolution? 

2.23 Differentiate between speed and velocity of a body. 

2.24 What is meant by retardation? 

2.25 What is the significance of velocity -time graph? 

2.26 Derive the equations of motion for an uniformly accelerated body. 

2.27 What are scalar and vector quantities? 

2.28 How will you represent a vector quantity? 

2.29 What is the magnitude and direction of the resultant of two 
vectors acting along the same line in the same direction? 

2.30 State: Parallelogram law of vectors and triangle law of vectors. 

2.31 Obtain the expression for magnitude and direction of the resultant 
of two vectors when they are inclined at an angle '0' with each 
other. 

2.32 State Newton's laws of motion. 

2.33 Explain the different types of inertia with examples. 

2.34 State and prove law of conservation of linear momentum. 

2.35 Define impulse of a force 

2.36 Obtain an expression for centripetal acceleration. 

2.37 What is centrifugal reaction? 



115 



2.38 Obtain an expression for the critical velocity of a body revolving 
in a vertical circle. 

2.39 What is meant by banking of tracks? 

2.40 Obtain an expression for the angle of lean when a cyclist takes a 
curved path. 

2.41 What are the two types of collision? Explain them. 

2.42 Obtain the expressions for the velocities of the two bodies after 
collision in the case of one dimensional motion. 

2.43 Prove that in the case of one dimensional elastic collision between 
two bodies of equal masses, they interchange their velocities 
after collision. 

Problems 

2.44 Determine the initial velocity and acceleration of particle travelling 
with uniform acceleration in a straight line if it travels 55 m in 
the 8 th second and 85 m in the 13 th second of its motion. 

2.45 An aeroplane takes off at an angle of 45° to the horizontal. If the 
vertical component of its velocity is 300 kmph, calculate its actual 
velocity. What is the horizontal component of velocity? 

2.46 A force is inclined at 60° to the horizontal . If the horizontal 
component of force is 40 kg wt, calculate the vertical component. 

2.47 A body is projected upwards with a velocity of 30 m s' 1 at an 
angle of 30° with the horizontal. Determine (a) the time of flight 
(b) the range of the body and (c) the maximum height attained by 
the body. 

2.48 The horizontal range of a projectile is 4^3 times its maximum 
height. Find the angle of projection. 

2.49 A body is projected at such an angle that the horizontal range is 
3 times the greatest height . Find the angle of projection. 

2.50 An elevator is required to lift a body of mass 65 kg. Find the 
acceleration of the elevator, which could cause a reaction of 
800 N on the floor. 

2.51 A body whose mass is 6 kg is acted on by a force which changes 
its velocity from 3 ms 1 to 5 m s' 1 . Find the impulse of the 

116 



force. If the force is acted for 2 seconds, find the force in 
newton. 

2.52 A cricket ball of mass 150 g moving at 36 m s' 1 strikes a bat and 
returns back along the same line at 21 m s' 1 . What is the 
change in momentum produced? If the bat remains in contact 
with the ball for 1 /20 s. what is the average force exerted in 
newton. 

2.53 Two forces of magnitude 12 N and 8 N are acting at a point. If 
the angle between the two forces is 60°, determine the magnitude 
of the resultant force? 

2.54 The sum of two forces inclined to each other at an angle is 
18 kg wt and their resultant which is perpendicular to the smaller 

force is 12 kg wt Find the forces and the angle between them. 

2.55 A weight of 20 kN supported by two cords, one 3 m long and the 
other 4m long with points of support 5 m apart. Find the tensions 
T 1 and T 2 in the cords. 

2.56 The following forces act at a point 

(i) 20 N inclined at 30° towards North of East 

(ii) 25 N towards North 

(Hi) 30 N inclined at 45° towards North of West 

(iv) 35 N inclined at 40° towards South of West. 

Find the magnitude and direction of the resultant force. 

2.57 Find the magnitude of the two forces such that it they are at 
right angles, their resultant is ^10 N. But if they act at 60°, their 
resultant is J13 N. 

2.58 At what angle must a railway track with a bend of radius 880 m 
be banked for the safe running of a train at a velocity of 
44ms- 1 ? 

2.59 A railway engine of mass 60 tonnes, is moving in an arc of 
radius 200 m with a velocity of 36 kmph. Find the force exerted 
on the rails towards the centre of the circle. 

2. 60 A horse pulling a cart exerts a steady horizontal pull of 300 N 

117 



and walks at the rate of 4.5 kmph. How much work is done by 
the horse in 5 minutes? 

2.61 A ball is thrown downward from a height of 30 m with a velocity 
of 10 m s' 1 . Determine the velocity with which the ball strikes 
the ground by using law of conservation of energy. 

2.62 What is the work done by a man in carrying a suitcase weighing 
30 kg over his head, when he travels a distance of 10 m in 
(i) vertical and (ii) horizontal directions? 

2.63 Two masses of 2 kg and 5 kg are moving with equal kinetic 
energies. Find the ratio of magnitudes of respective linear 
momenta. 

2.64 A man weighing 60 kg runs up a flight of stairs 3m high in 4 s. 
Calculate the power developed by him. 

2.65 A motor boat moves at a steady speed of 8 m s~ 1 , If the water 
resistance to the motion of the boat is 2000 N, calculate the 
power of the engine. 

2.66 Two blocks of mass 300 kg and 200 kg are moving toward each 
other along a horizontal frictionless surface with velocities of 50 
m s 1 and 100 m s' 1 respectively. Find the final velocity of each 
block if the collision is completely elastic. 



118 



Answers 

2.1 (c) 2.2 (c) 2.3 (d) 2.4 (c) 

2.5 (d) 2.6 (d) 2.7 (b) 2.8 (b) 

2.9 (b) 2.10 (d) 2.11 (b) 2.12 (c) 

2.13 (a) 2.14 (b) 2.15 (c) 2.16 (b) 

2.17 (c) 2.18 (a) 2.19 (d) 2.20 (c) 

2.44 10 ms- 1 ; 6 m s~ 2 2.45 424.26 kmph • 300 kmph 

2.46 69.28 kg wt 2.47 3.06s; 79.53 m; 11.48 m 

2.48 30° 2.49 53°7' 

2.50 2.5 ms 2 2.51 12 N s ; 6 N 

2.52 8.55 kgms- 1 ; 171 N 2.53 17.43 N 
2.54 5kg wt ; 13 kg wt ; 112°37' 2.55 16 kN, 12 kN 

2.56 45.6 N ; 132° 18' 2.57 3 N ; IN 

2.58 12°39' 2.59 30 kN 

2.60 1.125xl0 5 J 2.61 26.23 ms~ l 

2.62 2940 J ;0 2.63 0.6324 

2.64 441 W 2.65 16000 W 
2.66 -70 m s _i ; 80 m s _J 



119 



3. Dynamics of Rotational Motion 

3. 1 Centre of mass 

Every body is a collection of large number of tiny particles. In 
translatory motion of a body, every particle experiences equal displacement 
with time; therefore the motion of the whole body may be represented by 
a particle. But when the body rotates or vibrates during translatory 
motion, then its motion can be represented by a point on the body that 
moves in the same way as that of a single particle subjected to the same 
external forces would move. A point in the system at which whole mass of 
the body is supposed to be concentrated is called centre of mass of the 
body. Therefore, if a system contains two or more particles, its translatory 
motion can be described by the motion of the centre of mass of the system. 

3. 1. 1 Centre of mass of a two-particle system 

Let us consider a system consisting of two particles of masses m 1 
and m 2 . P, and P 2 are their positions at time t and r 1 and r 2 are the 
corresponding distances from the origin O as shown in Fig. 3. 1 . Then the 
velocity and acceleration of the particles are, 

dr. , , 

°'-dt - (1) 

do, 
a 1= ^ ...(2) 

dr 
v 2 = -f t ...(3) 

a 2 =-± ...(4) 




> X The particle at P 1 experiences two forces : 



O 

Fig 3.1 - Centre of mass (i) a force F J2 due to the particle at P 2 and 

(ii) force F Je , the external force due to some 
particles external to the system. 

If F l is the resultant of these two forces, 

120 



F = F + F 

r 1 r 12 r le 



.(5) 



Similarly, the net force F 2 acting on the particle P 2 is, 



F = F + F 

1 2 L 21 L 2e 



where F 21 is the force exerted by the particle at P 1 on P 2 
By using Newton's second law of motion, 



F 1 = m 1 a 1 
and F 2 = m 2 a 2 



.(6) 



■ (7) 
.(8) 



Adding equations (7) and (8), m 1 a 1 + m 2 a 2 = F 1 + F 2 
Substituting F T and F 2 from (5) and (6) 



m i a i + m 2 a 2 = F 12 + F le + F 21 + F 2e 



By Newton's third law, the internal force F 12 exerted by particle at 
P 2 on the particle at P l is equal and opposite to F 21 , the force exerted by 
particle at P, on P„. 



(i.e) F 



12 



* 21 

F + F 

r le r 2e 



...(9) 
...(10) 



[•.■ m l a l + m 2 a 2 = F] 

where F is the net external force acting on the system. 
The total mass of the system is given by, 



M = rrij+rT^ 



(11) 



Let the net external force F acting on the system produces an 
acceleration a CM called the acceleration of the centre of mass of the system 

By Newton's second law, for the system of two particles, 

F = Ma CM ...(12) 

From (10) and (12), M a CM = vn. 1 a 1 +m 2 a 2 ...(13) 

Let R CM be the position vector of the centre of mass. 



d A (R, 



CM) 



dt Z 

From (13) and (14), 



• (14) 



d 2 R c 



- + m, 



dt 2 



121 



2 A 



H 2 R 

dt 2 M ^dt 2 ' ! 



/ 



**CM 



U^h +m 2 r 2 ) 



m, r, +m,r, 

R„„ = — — ...(15) 

CM rrij +m 2 

This equation gives the position of the centre of mass of a system 
comprising two particles of masses m 1 and m^ 

If the masses are equal (rrij = m 2 ), then the position vector of the 
centre of mass is, 

r, +r ? 
R CM=^~ -(16) 

which means that the centre of mass lies exactly in the middle of the line 
joining the two masses. 

3. 1.2 Centre of mass of a body consisting ofn particles 

For a system consisting ofn particles with masses m,, m^, rru ... m n 
with position vectors r v r 2 , r 3 ...r n , the total mass of the system is, 

M = m 1 + m 2 +m 3 + +m n 

The position vector R„ M of the centre of mass with respect to origin 
O is given by 

n n 

Z m i r i Z m i r i 

_ m ; r ; +m 2 r 2 + m n r n = f^ = fg 

™ mi +m 2 + m n " M 

i=i 

The x coordinate and y coordinate of the centre of mass of the system 
are 

^ m I x j J^n g X3 + m„x n _ m^ + m 2 y 2 + m n y n 

x ~ 171^1712 + n^ and y - m] + m 2 + m„ 

Example for motion of centre of mass 

Let us consider the motion of the centre of mass of the Earth and 
moon system (Fig 3.2). The moon moves round the Earth in a circular 

122 



orbit and the Earth moves 
round the Sun in an elliptical 
orbit. It is more correct to say 
that the Earth and the moon 
both move in circular orbits 
about their common centre of 
mass in an elliptical orbit round 
the Sun. 




Centre of mass 



Fig 3.2 Centre of mass of Earth 
moon system 



For the system consisting 
of the Earth and the moon, their mutual gravitational attractions are the 
internal forces in the system and Sun's attraction on both the Earth and 
moon are the external forces acting on the centre of mass of the system. 

3.1.3Centre of gravity 

A body may be considered to be made up of an indefinitely large 
number of particles, each of which is attracted towards the centre of the 
Earth by the force of gravity. These forces constitute a system of like parallel 
forces. The resultant of these parallel forces known as the weight of the 
body always acts through a point, which is fixed relative to the body, 
whatever be the position of the body. This fixed point is called the centre 
of gravity of the body. 

The centre of gravity of a body is the point at which the resultant of 
the weights of all the particles of the body acts, whatever may be the 
orientation or position of the body provided that its size 
and shape remain unaltered. 

In the Fig. 3.3, W V W 2 ,W 3 are the weights 

of the first, second, third, ... particles in the body 
respectively. If W is the resultant weight of all 
the particles then the point at which W acts is 
known as the centre of gravity. The total weight 
of the body may be supposed to act at its centre 
of gravity. Since the weights of the particles 
constituting a body are practically proportional 
to their masses when the body is outside the 
Earth and near its surface, the centre of mass of 
Fig 3.3 Centre ofqravitu a body practically coincides with its centre of 

gravity. 

123 




w 3 w 4 



3.1.4 Equilibrium of bodies and types of equilibrium 

If a marble M is placed on a curved surface of a bowl S, it rolls down 
and settles in equilibrium at the lowest point A (Fig. 3.4 a}. This equilibrium 
position corresponds to minimum potential energy. If the marble is 
disturbed and displaced to a point B, its energy increases When it is 
released, the marble rolls back to A. Thus the marble at the position A is 
said to be in stable equilibrium. 

Suppose now that the bowl S is inverted and the marble is placed at 
its top point, at A (Fig. 3.4b). If the marble is displaced slightly to the 
point C, its potential energy is lowered and tends to move further away 
from the equilibrium position to one of lowest energy. Thus the marble is 
said to be in unstable equilibrium. 

Suppose now that the marble is 
placed on a plane surface (Fig. 3.4c). If 
it is displaced slightly, its potential 
energy does not change. Here the marble 
is said to be in neutral equilibrium. 




Unstable 




(b) 
Neutral 

£& Q. 



(c) 
Fig. 3. 4 Equilibrium of rigid bodies 



Equilibrium is thus stable, unstable 
or neutral according to whether the 
potential energy is minimum, maximum 
or constant. 

We may also characterize the 
stability of a mechanical system by 
noting that when the system is disturbed 
from its position of equilibrium, the 
forces acting on the system may 

(i) tend to bring back to its original 
position if potential energy is a 
minimum, corresponding to stable 
equilibrium. 



(ii) tend to move it farther away if potential energy is maximum, 
corresponding unstable equilibrium. 

(iii) tend to move either way if potential energy is a constant 
corresponding to neutral equilibrium 



124 



A 




B 



>Y 






e 



\ \ 
\ \ 
\ \ 
\ \ 

\ \ 

w v-' 



c 



/ N 
/ / 
/ / 

/ / 



\ \ 

\ 




*A 






w 




\ v 
\ \ 

\ \ 

\ N 

\ \ 

\ / 



(a) Stable equilibrium (b) Unstable equilibrium (c) Neutral equilibrium 

Fig 3.5 Types of equilibrium 

Consider three uniform bars shown in Fig. 3.5 a,b,c. Suppose each 
bar is slightly displaced from its position of equilibrium and then released. 
For bar A, fixed at its top end, its centre of gravity G rises to G 1 on being 
displaced, then the bar returns back to its original position on being 
released, so that the equilibrium is stable. 

For bar B, whose fixed end is at its bottom, its centre of gravity G is 
lowered to G 2 on being displaced, then the bar B will keep moving away 
from its original position on being released, and the equilibrium is said to 
be unstable. 

For bar C, whose fixed point is about its centre of gravity, the centre 
of gravity remains at the same height on being displaced, the bar will 
remain in its new position, on being released, and the equilibrium is said 
to be neutral. 

3.2 Rotational motion of rigid bodies 

3.2.1 Rigid body 

A rigid body is defined as that body which does not undergo any 
change in shape or volume when external forces are applied on it. When 
forces are applied on a rigid body, the distance between any two particles 
of the body will remain unchanged, however, large the forces may be. 

Actually, no body is perfectly rigid. Every body can be deformed 
more or less by the application of the external force. The solids, in which 
the changes produced by external forces are negligibly small, are usually 
considered as rigid body. 



125 



3.2.2 Rotational motion 



When a body rotates about a fixed axis, its motion is known as 
rotatory motion. A rigid body is said to have pure rotational motion, if every 
particle of the body moves in a circle, the centre of which lies on a straight 
line called the axis of rotation (Fig. 3.6). The axis of rotation may lie inside 
the body or even outside the body. The particles lying 
on the axis of rotation remains stationary. 

The position of particles moving in a circular path 
is conveniently described in terms of a radius vector r 
and its angular displacement . Let us consider a rigid 
body that rotates about a fixed axis XOX' passing 
through O and perpendicular to the plane of the paper 
as shown in Fig 3.7. Let the body rotate from the position 
A to the position B. The different particles at P 1 ,P 2 ,P 3 . 
.... in the rigid body covers unequal distances PjP{, J^Y' 
„, PgPq'.... in the same interval of 
time. Thus their linear 
velocities are different. But in 
the same time interval, they all rotate through 
the same angle 6 and hence the angular velocity 
is the same for the all the particles of the rigid 
body. Thus, in the case of rotational motion, 
different constituent particles have different linear 
j B velocities but all of them have the same angular 
velocity. 





Axis of rotation 

Fig 3.6 Rotational 

motion 



\ 



3.2.3 



Equations of rotational motion 



Fig 3. 7 Rotational 
motion of a rigid body 



As in linear motion, for a body having 
uniform angular acceleration, we shall derive the 
equations of motion. 



Let us consider a particle start rotating with angular velocity ra and 
angular acceleration a. At any instant t, let co be the angular velocity of 
the particle and 6 be the angular displacement produced by the particle. 

Therefore change in angular velocity in time t= co - co 

change in angular velocity 



But, angular acceleration 



time taken 



126 



(i.e) a = —^ -(I) 

co = co +at •••(2) 



The average angular velocity 



a> + a>„ 



2 



The total angular displacement 

= average angular velocity x time taken 



m + co. 
(i.e) e = I —^ 



t ...(3) 



( m + at + co a 
Substituting co from equation (2), 8 = ; 

1 , 
6 = ra o t+ yat 2 ...(4) 

(co- CO, ") 
From equation (1), t = ...(5) 

using equation (5) in (3), 

f co + a>, "| ( co-coA [a 1 -co]) 
6= l 2 j { a ) =^^r~ 

2a 9 = ca 2 - co 2 or co 2 = cog + 2a 6 . . .(6) 

Equations (2), (4) and (6) are the equations of rotational motion. 

3.3 Moment of inertia and its physical significance 

According to Newton's first law of motion, a body must continue in 
its state of rest or of uniform motion unless it is compelled by some 
external agency called force. The inability of a material body to change 
its state of rest or of uniform motion by itself is called inertia. Inertia is 
the fundamental property of the matter. For a given force, the greater the 
mass, the higher will be the opposition for motion, or larger the inertia. 
Thus, in translatory motion, the mass of the body measures the coefficient 
of inertia. 

Similarly, in rotational motion also, a body, which is free to rotate 
about a given axis, opposes any change desired to be produced in its 
state. The measure of opposition will depend on the mass of the body 

127 



and the distribution of mass about the axis of rotation. The coefficient of 
inertia in rotational motion is called the moment of inertia of the body 
about the given axis. 

Moment of inertia plays the same role in rotational motion as that 
of mass in translatory motion. Also, to bring about a change in the state 
of rotation, torque has to be applied. 

3.3.1 Rotational kinetic energy and moment of inertia of a rigid 
body 

Consider a rigid body rotating with angular velocity co about an axis 
XOX'. Consider the particles of masses m 1 , m 2 , m 3 ... situated at distances 
r 1 r 2 , r 3 ... respectively from the axis of rotation. The angular velocity of all 
the particles is same but the particles rotate with different linear velocities. 
Let the linear velocities of the particles be v 1 ,v 2 ,v 3 ... respectively. 

Kinetic energy of the first particle = - m^ 2 

But u i =r J co 

.*. Kinetic energy of the first particle 

= rm/ rjco) 2 = - nr,^ 2 © 2 

Similarly, 

Kinetic energy of second particle 

1 



Kinetic energy of third particle 
= y m 3 r 3 2 ® 2 and so on. 




Fig. 3.8 Rotational kinetic energy 
and moment of inertia 



The kinetic energy of the rotating rigid body is equal to the sum of 
the kinetic energies of all the particles. 

.". Rotational kinetic energy 

= — (m 1 r 1 2 <o 2 + m 2 r 2 2 (d 2 + m 3 r 3 2 co 2 + + m n r n 2 irP) 

= - to 2 (m i r I 2 + m 2 r 2 + m 3 r 2 + + m n r n 2 ) 



128 



(i.e) E R =\a 2 [t m ^ 



-(I) 



In translatory motion, kinetic energy = — mv 2 

Comparing with the above equation, the inertial role is played by 

n 

the term z_, m \ r i . This is known as moment of inertia of the rotating rigid 

i =1 

body about the axis of rotation. Therefore the moment of inertia is 
I = mass x (distance ) 2 

Kinetic energy of rotation = — co 2 I 

When co = 1 rad s" 1 , rotational kinetic energy 

= E R = \ (1) 2 I (or) I=2E R 

It shows that moment of inertia of a body is equal to twice the kinetic 
energy of a rotating body whose angular velocity is one radian per second. 

The unit for moment of inertia is kg m 2 and the dimensional formula 
is ML 2 . 

3.3.2 Radius of gyration 

The moment of inertia of the rotating rigid body is, 

n 

1= S mf 2 = m 1 r 2 ^ m 2 r 2 + ...m n r 2 
If the particles of the rigid body are having same mass, then 

m i = m 2 = m 3 = = m ( sa y) 

.". The above equation becomes, 

I = mr 1 2 + mr 2 2 + mr 3 2 + + mr n 2 

= m (r 2 + r 2 2 + r 3 2 + + r 2 ) 

I = nm 



r i 2 + r 2 Z + r 3 2 ' 



where n is the number of particles in the rigid body. 

129 



I = MK 2 ... (2) 



where M = nm, total mass of the body and K 



2 . 



ri 2 +r 2 2 +r 3 2 + r n 2 



k 2 +r 2 2 +r 3 2 + r n 2 

Here K= J is called as the radius of gyration of the 

V n 

rigid body about the axis of rotation. 

The radius of gyration is equal to the root mean square distances of 
the particles from the axis of rotation of the body. 

The radius of gyration can also be defined as the perpendicular 
distance between the axis of rotation and the point where the whole weight 
of the body is to be concentrated. 

Also from the equation (2) K 2 = — (or) K = J— 

3.3.3 Theorems of moment of inertia 
(i) Parallel axes theorem 

Statement 

The moment of inertia of a body about any axis is equal to the sum of 
its moment of inertia about a parallel axis through its centre of gravity and 
the product of the mass of the body and the square of the distance between 
the two axes. 

Proof 

Let us consider a body having its centre of gravity at G as shown in 
Fig. 3.9. The axis XX' passes through the centre of gravity and is 
perpendicular to the plane of the body. The axis X 1 X 1 ' passes through 
the point O and is parallel to the axis XX' . The distance between the two 
parallel axes is x. 

Let the body be divided into large number of particles each of mass 
m . For a particle P at a distance r from O, its moment of inertia about the 
axis XjOXj' is equal to mr 2 . 

The moment of inertia of the whole body about the axis X 1 X 1 ' is 
given by, 

I =Zmr 2 ...(1) 

130 



From the point P, drop a perpendicular PA to the extended OG and 
join PG. 

iX 




x/ : 

Fig .3.9 Parallel axes theorem 



In the AOPA, 

OP 2 =OA 2 + AP 2 

r 2 = (x+h) 2 +AP 2 

r 2 = x 2 +2xh + h 2 + AP 2 ...(2) 

But from A GPA, 
GP 2 = GA 2 + AP 2 
y 2 = h 2 + AP 2 

Substituting equation (3) in (2), 
r 2 = x 2 + 2xh + y 2 

Substituting equation (4) in (1), 
I = Zm(x 2 + 2xh + y 2 ) 
= Zmx 2 + Z2mxh + Zmy 2 
= Mx 2 + My 2 + 2xZmh 



.(3) 



...(4) 



...(5) 



Here My 2 = I Q is the moment of inertia of the body about the line 
passing through the centre of gravity. The sum of the turning moments of 



131 



all the particles about the centre of gravity is zero, since the body is 
balanced about the centre of gravity G. 

E (mg) (h) = (or) E mh = [since g is a constant] ■••(6) 

..equation (5) becomes, I Q = Mx 2 + 1„ ••■(7) 

Thus the parallel axes theorem is proved. 

(ii) Perpendicular axes theorem 

Statement 

The moment of inertia of a plane laminar body about an axis 
perpendicular to the plane is equal to the sum of the moments of inertia 
about two mutually perpendicular axes in the plane of the lamina such that 
the three mutually perpendicular axes have a common point of intersection. 

Proof 

Consider a plane lamina having 
the axes OX and OY in the plane of the 
lamina as shown Fig. 3.10. The axis 
OZ passes through O and is 
perpendicular to the plane of the 
lamina. Let the lamina be divided into 
a large number of particles, each of 
mass m. A particle at P at a distance r 
from O has coordinates (x,y) . Fig 3.10 Perpendicular axes theorem 

;.r 2 =x 2 +y 2 ...(1) 

The moment of inertia of the particle P about the axis OZ = mr 2 . 
The moment of inertia of the whole lamina about the axis OZ is 

I z = Zmr 2 ...(2) 

The moment of inertia of the whole lamina about the axis OX is 

I x =Zmy 2 ...(3) 




Similarly, J = Z mx 2 
From eqn. (2), I z = Emr 2 = Em(x 2 ^y 2 ) 
I z =Zmx 2 +Zmy 2 = I y + I x 

which proves the perpendicular axes theorem. 



.(4) 



132 



Table 3. 1 Moment of Inertia of different bodies 

(Proof is given in the annexure) 



Body 


Axis of Rotation 


Moment of Inertia 


Thin Uniform Rod 


Axis passing through its 
centre of gravity and 
perpendicular to its length 


Ml 2 M - mass 
12 l - length 




Axis passing through the 
end and perpendicular to 
its length. 


Ml M - mass 
3 1 - length 


Thin Circular Ring 


Axis passing through its 
centre and perpendicular 
to its plane. 


MR 2 M ~ mass 
R - radius 




Axis passing through its 
diameter 


1 2 M-mass 

2 R - radius 




Axis passing through a 
tangent 


^MR 2 M-mass 
2 R - radius 


Circular Disc 


Axis passing through its 
centre and perpendicular 
to its plane. 


-MR 2 M " mass 
2 R - radius 




Axis passing through its 
diameter 


1 M - mass 
4 MR R - radius 




Axis passing through a 
tangent 


- MR 2 M " mass 
4 R - radius 


Solid Sphere 


Axis passing through its 
diameter 


2 2 M " mass 
5^ R - radius 




Axis passing through a 
tangent 


7 M - mass 
- MR 2 R . radius 


Solid Cylinder 


Its own axis 


1 M - mass 
2 MR R- radius 




Axis passing through its 
centre and perpedicular to 
its length 


(rt i 2 ) 


M - mass 
R - radius 
1 - length 



133 



3.4 Moment of a force 

A force can rotate a nut when applied by a wrench or it can open a 
door while the door rotates on its hinges (i.e) in addition to the tendency 
to move a body in the direction of the application of a force, a force also 
tends to rotate the body about any axis which does not intersect the line 
of action of the force and also not parallel to it. This tendency of rotation 
is called turning effect of a force or moment of the force about the given 
axis. The magnitude of the moment of force F about a point is defined as 
the product of the magnitude of force and the perpendicular distance of the 
point from the line of action of the force. A^g 

Let us consider a force Facting at the 
point P on the body as shown in Fig. 3.11. 
Then, the moment of the force F about the 
point O = Magnitude of the force x 
perpendicular distance between the 
direction of the force and the point about 
which moment is to be determined = F x OA. 

If the force acting on a body rotates Fi 9 3U Moment °J~ a J orce 

the body in anticlockwise direction with respect to O then the moment is 
called anticlockwise moment. On the other hand, if the force rotates the 

body in clockwise direction then the moment 
^F : is said to be clockwise moment. The unit of 
moment of the force is N m and its 
dimensional formula is M L 2 T 2 . 




o 



O' 



->. 



Fig 3.12 Clockwise and 
anticlockwise moments 



As a matter of convention,an anticlockwise 
moment is taken as positive and a clockwise 
&F 2 moment as negative. While adding moments, 
the direction of each moment should be taken 
into account. 



In terms of vector product, the moment of a force is expressed as, 
m = r x F 

where r is the position vector with respect to O. The direction of m is 
perpendicular to the plane containing r and F. 



134 




F Fig. 3.13 Couple 



3.5 Couple and moment of the couple (Torque) 

There are many examples in practice where 
two forces, acting together, exert a moment, or 
turning effect on some object. As a very simple 
case, suppose two strings are tied to a wheel at 
the points X and Y, and two equal and opposite x 
forces, F, are exerted tangentially to the wheels 
(Fig. 3. 13). If the wheel is pivoted at its centre O 
it begins to rotate about O in an anticlockwise 
direction. 

Two equal and opposite forces whose lines 
of action do not coincide are said to constitute a couple in mechanics. The 
two forces always have a turning effect, or moment, called a torque. The 
perpendicular distance between the lines of action of two forces, which 
constitute the couple, is called the arm of the couple. 

The product of the forces forming the couple and the arm of the couple 
is called the moment of the couple or torque. 

Torque = one of the forces x perpendicular distance between the 
forces 

The torque in rotational motion plays the same role as the force in 
translational motion. A quantity that is a measure of this rotational effect 
produced by the force is called torque. 

In vector notation, r = r x F 

The torque is maximum when 8 = 90° (i.e) when the applied force is 
at right angles to r . 

Examples of couple are 

1 . Forces applied to the handle of a 
screw press, 

2. Opening or closing a water tap. 

3. Turning the cap of a pen. 

4. Steering a car. 

Work done by a couple 

Suppose two equal and opposite forces F act tangentially to a wheel 
W, and rotate it through an angle 6 (Fig. 3.14). 




Fig.3.14 Work done by a 
couple 



135 



»x 



Then the work done by each force = Force x distance = F x r 6 
(since r 6 is the distance moved by a point on the rim) 

Total work done W = Fr8 + Fr8 = 2FrO 

but torque r = F x 2r = IF r 

.'. work done by the couple, W = r 

3.6 Angular momentum of a particle 

The angular momentum in a rotational motion is similar to the 
linear momentum in translatory motion. The linear momentum of a 
particle moving along a straight line is the z 

product of its mass and linear velocity (i.e) p = 
mv. The angular momentum of a particle is 
defined as the moment of linear momentum of 
the particle. 

Let us consider a particle of mass m / ^\_p^S-^p/ 

moving in the XY plane with a velocity v and / j^/ 

linear momentum p =mv at a distance r from y 

the origin (Fig. 3.15). Y Fi 9 3.15 Angular 

momentum of a particle 

The angular momentum L of the particle 

about an axis passing through O perpendicular to XY plane is defined as 
the cross product of r and p . 

(i.e) L =r xp 

Its magnitude is given by L = r p sin 

where 8 is the angle between r and p and L is along a direction 
perpendicular to the plane containing r and p . 

The unit of angular momentum is kg m 2 s _1 and its dimensional 
formula is, ML 2 r'. 

3.6.1 Angular momentum of a rigid body 

Let us consider a system of n particles of masses m,, m^ m n 

situated at distances r v r 2 r n respectively from the axis of rotation 

(Fig. 3.16). Let v lt v 2 , i> 3 be the linear velocities of the particles 

respectively, then linear momentum of first particle = m 1 v l . 



136 




Fig 3.16 Angular momentum of a 
rigid body 



Since u 1 = ^co the linear momentum 
of first particle = m 1 [r l co) 

The moment of linear 
momentum of first particle 

= linear momentum x 

perpendicular distance 

= (m^co) x r 1 

angular momentum of first 
particle = m 1 r 1 2 (o 

Similarly, 

angular momentum of second particle = n^r^co 

angular momentum of third particle = m^r^m and so on. 

The sum of the moment of the linear momenta of all the particles of a 
rotating rigid body taken together about the axis of rotation is known as 
angular momentum of the rigid body. 

:. Angular momentum of the rotating rigid body = sum of the angular 
momenta of all the particles. 



m 1 r 1 2 © + m 2 r^-& + m^r^ca 



+ rry^co 



L = co [m 1 r 1 + mjr 2 + rn^r 2 + m n r n 2 ] 



Urn^ 2 



L=cal 

n 



where I = 2—i ™~fi = moment of inertia of the rotating rigid body about 



i=l 

the axis of rotation. 



3.7 Relation between torque and angular acceleration 

Let us consider a rigid body rotating about a fixed axis XOX' with 
angular velocity co (Fig. 3. 17). 

The force acting on a particle of mass rrij situated at A, at a distance 
r,, from the axis of rotation = mass x acceleration 

d < i 
= m. x — llfflj 

1 dt 



137 



= m,r 1 ^_ 
dt 

dt 2 
The moment of this force 
about the axis of rotation 

= Force x perpendicular distance 
d 2 9 

■ m ^ W x r i 

Therefore, the total moment of all 
the forces acting on all the particles 




Fig 3.17 Relation between torque and 
angular acceleration 



m,r 



2 d 2 e 2 d 2 e 



V 1 



dt 2 



+ m n r. 



2' 2 



dt 2 



+ ... 



n d 2 B 

(i.e) torque = Z m ^ x ~^T 



or 



t = la 



where 2-i t i = moment of inertia I of the rigid body and a = — — angular 

(=i dt 

acceleration. 

3.7.1 Relation between torque and angular momentum 

The angular momentum of a rotating rigid body is, L = I co 
Differentiating the above equation with respect to time, 

dt {dt) 

dco 
where a = — — angular acceleration of the body. 



But torque r = la 
Therefore, torque x = 



dL 
~dt 



Thus the rate of change of angular momentum of a body is equal to 
the external torque acting upon the body. 

3.8 Conservation of angular momentum 

The angular momentum of a rotating rigid body is, L = I co 

dL 



The torque acting on a rigid body is, x = 



dt 



138 



dL 
~dt 







When no external torque acts on the system, x 

(i.e) L = I co = constant 

Total angular momentum of the body = constant 

(i.e.) when no external torque acts on the body, the net angular 
momentum of a rotating rigid body remains constant. This is known as law 
of conservation of angular momentum. 

Illustration of conservation of angular momentum 

From the law of conservation of angular momentum, Joo = constant 

1 

(ie) co x — , the angular velocity of rotation is inversely proportional 

to the moment of inertia of the system. 

Following are the examples for law of conservation of angular 
momentum. 

1 . A diver jumping from springboard sometimes exhibits somersaults 
in air before reaching the water surface, because the diver curls his body 
to decrease the moment of inertia and increase angular velocity. When he 




Fig. 3.18 A diver jumping from a spring board 



139 



is about to reach the water surface, he again outstretches his limbs. This 
again increases moment of inertia and decreases the angular velocity. 
Hence, the diver enters the water surface with a gentle speed. 

2. A ballet dancer can increase her angular velocity by folding her 
arms, as this decreases the moment of inertia. 








wit 




Fig 3.19 A person rotating on a turn table 

3. Fig. 3. 19a shows a person sitting on a turntable holding a pair of 
heavy dumbbells one in each hand with arms outstretched. The table is 
rotating with a certain angular velocity. The person suddenly pushes the 
weight towards his chest as shown Fig. 3.19b, the speed of rotation is 
found to increase considerably. 

4.The angular velocity of a planet in its orbit round the sun increases 
when it is nearer to the Sun, as the moment of inertia of the planet about 
the Sun decreases. 



140 



Solved Problems 



3.1 A system consisting of two masses connected by a massless rod 
lies along the X-axis. A 0.4 kg mass is at a distance x = 2 m while 
a 0.6 kg mass is at x = 7 m. Find the x coordinate of the centre 
of mass. 
Data : m 1 = 0.4 kg ; m 2 = 0.6 kg ; x x =2m;x 2 = 7m;x=? 



Solution 



x 



m lXl 



m 1 x 2 (0.4x2) + (0.6x7) 



m, 



m, 



(0.4 + 0.6) 



-5 m 



3.2 Locate the centre of mass of a system of bodies of masses 
m 1 = 1 kg, m 2 = 2 kg and m 3 = 3 kg situated at the corners of 
an equilateral triangle of side 1 m. 

Data : m 1 = 1 kg ; m 2 = 2 kg ; m 3 = 3 kg ; 

The coordinates of A = (0,0) 

The coordinates of B =(1,0) 

Centre of mass of the system =? 

Solution : Consider an equilateral triangle of side lmas shown 

in Fig. Take X and Y axes as 
shown in figure. 

To find the coordinate of C: 

For an equilateral triangle , 
ZCAB = 60° 

Consider the triangle ADC, 

. „ CD 



CA 



(or) CD 




> X 



(CA) sine = lxsin60° 



43 



Therefore from the figure, the coordinate of C are, ( 0.5, J 

_ m 1 x 1 + m 2 x 2 + m 3 x 3 



m l +m 2 + m 3 



141 



(Ix0) + (2xl) + (3x0.5) 3.5 

x = = m 

(1 + 2 + 3) 6 



m 1 y 1 + m 2 y 2 + nyy^ 



m 1 + m 2 +m 3 

3 



( 
(Ix0) + (2x0) + 



3 x 

v 2 J >/3 
u = = m 

6 4 



3.3 A circular disc of mass m and radius r is set rolling on a table. 
If co is its angular velocity, show that its total energy E = — mr 2 oo 2 . 

Solution : The total energy of the disc = Rotational KE + linear KE 

1 9 1 9 

.-. E =- Ia 2 + — mv 2 ...(1) 

But I = — mr 2 and u = rco ...(2) 

Substituting eqn. (2) in eqn. (1), 

E = — x — ( mr 2 ) (co 2 )+ — m (rco) 2 = — mr 2 oi 2 + — mr 2 a> 2 
2 2 2 4 2 

3 2 2 

4 

3.4 A thin metal ring of diameter 0.6m and mass 1kg starts from 
rest and rolls down on an inclined plane. Its linear velocity on 
reaching the foot of the plane is 5 m s" 1 , calculate (i) the moment 
of inertia of the ring and (ii) the kinetic energy of rotation at that 
instant. 

Data : R = 0.3 m ; M = 1 kg ; v = 5 m s _J ; I =? K.E. = ? 

Solution : I = MR 2 = 1 x (0.3) 2 = 0.09 kg m 2 

1 
K.E. 



2*° 2 










■ :. oo = 


V 
r 


K.E. -- 


1 
= - x 0.09 x 


I 0.3 



v = m ; :. oo = - ; K.E. = — x 0.09 x \ ^^7 = 12.5 J 

142 



3.5 A solid cylinder of mass 200 kg rotates about its axis with angular 
speed 100 s" 1 . The radius of the cylinder is 0.25 m. What is the 
kinetic energy associated with the rotation of the cylinder? What 
is the magnitude of the angular momentum of the cylinder about 
its axis? 

Data : M = 200 kg ; a> = 100 s' 1 •' R = 0.25 metre ; 



E R =?;L = ? 



MR 2 200x(0.25) 2 



Solution : I = = = 6.25 kq m z 

2 2 

1 „ 
K.E. = —la 2 

= — x 6.25 x (100) 2 

E R = 3.125 x 10 4 J 

L =103= 6.25 x 100 = 625 kg m 2 s _i 

3.6 Calculate the radius of gyration of a rod of mass 100 g and length 
100 cm about an axis passing through its centre of gravity and 
perpendicular to its length. 

Data : M = 100 g = 0.1 kg I = 100 cm = 1 m 

K = ? 

Solution : The moment of inertia of the rod about an axis passing 
through its centre of gravity and perpendicular to the length = I = 

ML 2 r 2 L 1 

MK 2 = -J2 (or) K 2 = ± (or) K = -^ = ^ = 0.2886 m. 

3.7 A circular disc of mass 100 g and radius 10 cm is making 2 
revolutions per second about an axis passing through its centre 
and perpendicular to its plane. Calculate its kinetic energy. 

Data : M = 100 g = 0.1 kg ; R = 10 cm = 0.1 m ; n = 2 

Solution ; © = angular velocity = 2nn = 2n x 2 = An rad / s 

1 9 

Kinetic energy of rotation = — Ico^ 

= — x — x MR 2 m 2 = — x — (0.1) x (0.1) 2 x (4k) 2 
= 3.947 x 10~ 2 J 

143 



3.8 Starting from rest, the flywheel of a motor attains an angular 
velocity 100 rad/s from rest in 10 s. Calculate (i) angular 
acceleration and (ii) angular displacement in 10 seconds. 

Data : a> = ; a> = 100 rad s _i t = 10 s a = ? 

Solution : From equations of rotational dynamics, 

a> = a> + at 

, . 0)-Q} n 100-0 , _ j -2 

(or) a = ^_ = = xo rad s z 

t 10 

Angular displacement 6 = a> Q t + —at 2 

= + — x 10 x 10 2 = 500 rad 
2 

3.9 A disc of radius 5 cm has moment of inertia of 0.02 kg m 2 .A force 
of 20 N is applied tangentially to the surface of the disc. Find the 
angular acceleration produced. 

Data : I = 0.02 kg m 2 ; r = 5 cm = 5 x lCr 2 m ; F = 20 N ; r = ? 
Solution : Torque = x = F x 2r = 20 x 2 x 5 x 10~ 2 = 2 N m 

r 2 9 

anqular acceleration = a = — = ^—~- =100 rad I s 

I 0.02 

3.10 From the figure, find the moment of the force 45 N about A? 

45 N 




,30° 



m m — w 

6m 2m 



Data : Force F = 45 N ; Moment of the force about A = ? 

Solution : Moment of the force about A 45 n 

= Force x perpendicular 6m 

distance = F x AO 

= 45 x 6 sin 30 = 135 N m 



\ 3rf^ „_ 

\ s 2m 

>6' 




,3C° 



144 



Self evaluation 

(The questions and problems given in this self evaluation are only samples. 
In the same way any question and problem could be framed from the text 
matter. Students must be prepared to answer any question and problem 
from the text matter, not only from the self evaluation.) 

3.1 The angular speed of minute arm in a watch is : 
(a) k l '21600 rad s" J (b) k/ 12 rad s" 2 
(c) k l '3600 rad s _i (d) n 1 1800 rad s _i 

3.2 The moment of inertia of a body comes into play 

(a) in linear motion (b) in rotational motion 

(c) in projectile motion (d) in periodic motion 

3.3 Rotational analogue of mass in linear motion is 

(a) Weight (b) Moment of inertia 

(c) Torque (d) Angular momentum 

3.4 The moment of inertia of a body does not depend on 

(a) the angular velocity of the body 

(b) the mass of the body 

(c) the axis of rotation of the body 

(d) the distribution of mass in the body 

3.5 A ring of radius r and mass m rotates about an axis passing 
through its centre and perpendicular to its plane with angular 
velocity co. Its kinetic energy is 

(a) mm 2 (b) — mrco 2 (c) lco 2 (d) — lco 2 

2 -^ 

3.6 The moment of inertia of a disc having mass M and radius R, 
about an axis passing through its centre and perpendicular to its 
plane is 

(a) - MR 2 (b) MR 2 (c) - MR 2 (d) - MR 2 

3.7 Angular momentum is the vector product of 

(a) linear momentum and radius vector 

(b) moment of inertia and angular velocity 

(c) linear momentum and angular velocity 

(d) linear velocity and radius vector 

145 



3.8 The rate of change of angular momentum is equal to 

(a) Force (b) Angular acceleration 

(c) Torque (d) Moment of Inertia 

3.9 Angular momentum of the body is conserved 

(a) always 

(b) never 

(c) in the absence of external torque 

(d) in the presence of external torque 

3.10 A man is sitting on a rotating stool with his arms outstretched. 
Suddenly he folds his arm. The angular velocity 

(a) decreases (b) increases 

(c) becomes zero (d) remains constant 

3.11 An athlete diving off a high springboard can perform a variety 
of exercises in the air before entering the water below. Which 
one of the following parameters will remain constant during the 
fall. The athlete's 

(a) linear momentum (b) moment of inertia 

(c) kinetic energy (d) angular momentum 

3.12 Obtain an expression for position of centre of mass of two particle 
system. 

3.13 Explain the motion of centre of mass of a system with an example. 

3.14 What are the different types of equilibrium? 

3.15 Derive the equations of rotational motion. 

3.16 Compare linear motion with rotational motion. 

3.17 Explain the physical significance of moment of inertia. 

3.18 Show that the moment of inertia of a rigid body is twice the 
kinetic energy of rotation. 

3.19 State and prove parallel axes theorem and perpendicular axes 
theorem. 

3.20 Obtain the expressions for moment of inertia of a ring (i) about 
an axis passing through its centre and perpendicular to its plane, 
(ii) about its diameter and (Hi) about a tangent. 

146 



3.21 Obtain the expressions for the moment of inertia of a circular 
disc (i) about an axis passing through its centre and perpendicular 
to its plane. (ii) about a diameter (Hi) about a tangent in its plane 
and (iv) about a tangent perpendicular to its plane. 

3.22 Obtain an expression for the angular momentum of a rotating 
rigid body. 

3.23 State the law of conservation of angular momentum. 

3.24 A cat is able to land on its feet after a fall. Which principle of 
physics is being used? Explain. 

Problems 

3.25 A person weighing 45 kg sits on one end of a seasaw while a 
boy of 15 kg sits on the other end. If they are separated by 
4 m, how far from the boy is the centre of mass situated. Neglect 
weight of the seasaw. 

3.26 Three bodies of masses 2 kg, 4 kg and 6 kg are located at the 
vertices of an equilateral triangle of side 0.5 m. Find the centre 
of mass of this collection, giving its coordinates in terms of a 
system with its origin at the 2 kg body and with the 4 kg body 
located along the positive X axis. 

3.27 Four bodies of masses 1 kg, 2 kg, 3 kg and 4 kg are at the 
vertices of a rectangle of sides a and b. If a = 1 m and 
b = 2 m, find the location of the centre of mass. (Assume that, 
1 kg mass is at the origin of the system, 2 kg body is situated 
along the positive x axis and 4 kg along the y axis.) 

3.28 Assuming a dumbbell shape for the carbon monoxide (CO) 
molecule, find the distance of the centre of mass of the molecule 
from the carbon atom in terms of the distance d between the 
carbon and the oxygen atom. The atomic mass of carbon is 
12 amu and for oxygen is 16 amu. (1 amu = 1.67 x 10 ~ 27 kg) 

3.29 A solid sphere of mass 50 g and diameter 2 cm rolls without 
sliding with a uniform velocity of 5 m s' 1 along a straight line on 
a smooth horizontal table. Calculate its total kinetic energy. 

( Note : Total E K = - mv 2 + -Jco 2 J. 



147 



3.30 Compute the rotational kinetic energy of a 2 kg wheel rotating at 
6 revolutions per second if the radius of gyration of the wheel is 
0.22 m. 

3.31 The cover of a jar has a diameter of 8 cm. Two equal, but 
oppositely directed, forces of 20 N act parallel to the rim of the 
lid to turn it. What is the magnitude of the applied torque? 



Answers 

3.1 (d) 3.2 (b) 3.3 (b) 3.4 (a) 

3.5 (d) 3.6 (a) 3.7 (b) 3.8 (c) 

3.9(c) 3.10 (b) 3.11 (c) 

3.25 3 mfrom the boy 3.26 0.2916 m, 0.2165 m 

16 d 
3.27 0.5 m, 1.4 m 3.28 -—- 

zo 

3.29 0.875 J 3.30 68.71 J 

3.31 1.6 N m 



148 



4. Gravitation and Space Science 



We have briefly discussed the kinematics of a freely falling body 
under the gravity of the Earth in earlier units. The fundamental forces 
of nature are gravitational, electromagnetic and nuclear forces. The 
gravitational force is the weakest among them. But this force plays an 
important role in the birth of a star, controlling the orbits of planets and 
evolution of the whole universe. 

Before the seventeenth century, scientists believed that objects fell 
on the Earth due to their inherent property of matter. Galileo made a 
systematic study of freely falling bodies. 

4. 1 Newton's law of gravitation 

The motion of the planets, the moon and the Sun was the interesting 
subject among the students of Trinity college at Cambridge in England. 
Moon Isaac Newton was also one among these 
students. In 1665, the college was closed for an 
indefinite period due to plague. Newton, who 
was then 23 years old, went home to 
Lincolnshire. He continued to think about the 
motion of planets and the moon. One day 
Newton sat under an apple tree and had tea 
with his friends. He saw an apple falling to 
ground. This incident made him to think about 
falling bodies. He concluded that the same force 
of gravitation which attracts the apple to the 
Earth might also be responsible for attracting 
the moon and keeping it in its orbit. The centripetal acceleration of the 
moon in its orbit and the downward acceleration of a body falling on the 
Earth might have the same origin. Newton calculated the centripetal 
acceleration by assuming moon's orbit (Fig. 4.1) to be circular. 

Acceleration due to gravity on the Earth's surface, g = 9.8 m s~ 2 

2 

Centripetal acceleration on the moon, a = — 




Fig. 4.1 Acceleration 
of moon 



149 



where r Is the radius of the orbit of the moon (3.84 x 10 8 m) and v is 
the speed of the moon. 

Time period of revolution of the moon around the Earth, 

T = 27.3 days. 

2;rr 
The speed of the moon in its orbit, v = — — 

2;rx3.84xl0 8 , M , ~ , 

v = = 1.02 x 10 3 ms- 1 



27.3x24x60x60 
Centripetal acceleration, a = 



v 2 (1.02xl0 3 ) 2 



3.84xlO a 



a = 2.7 x 10 



■3 



Newton assumed that both the moon and the apple are accelerated 
towards the centre of the Earth. But their motions differ, because, the 
moon has a tangential velocity whereas the apple does not have. 

Newton found that a c was less than g and hence concluded that 
force produced due to gravitational attraction of the Earth decreases 
with increase in distance from the centre of the Earth. He assumed that 
this acceleration and therefore force was inversely proportional to the 
square of the distance from the centre of the Earth. He had found that 
the value of a c was about 1/3600 of the value of g, since the radius of 
the lunar orbit r is nearly 60 times the radius of the Earth R. 

The value of a c was calculated as follows : 

,2 1 



a c 


.l/r 2 _(R) 2 _( 1 


9 


l/R 2 [rj {60 




9 _ 9.8 



3600 



" c 3600 3600 

Newton suggested that gravitational force might vary inversely as 
the square of the distance between the bodies. He realised that this 
force of attraction was a case of universal attraction between any two 
bodies present anywhere in the universe and proposed universal 
gravitational law. 

The law states that every particle of matter in the universe attracts 
every other particle with a force which is directly proportional to the product 



150 



of their masses and inversely proportional to the square of the distance 
between them. 

Consider two bodies of masses m 1 and m^ with their centres 
separated by a distance r. The gravitational force between them is 



Fa m.^2 



Fa 1/r 2 



m, 



F a 2 

r 



m^ \^7 r 




Fig. 4.2 
Gravitational 
F = G — 2 — where G is the universal force 



m^j 



r 
gravitational constant. 

If m 1 = m 2 = 1 kg and r = 1 m, then F = G. 

Hence, the Gravitational constant 'G' is numerically equal to 
the gravitational force of attraction between two bodies of mass 
1 kg each separated by a distance of 1 m. The value of G is 
6.67 x 10~ u N m 2 kg 2 and its dimensional formula is M" 1 L 3 T~ 2 . 

4.1.1 Special features of the law 

(i) The gravitational force between two bodies is an action and 
reaction pair. 

(ii) The gravitational force is very small in the case of lighter 
bodies. It is appreciable in the case of massive bodies. The gravitational 
force between the Sun and the Earth is of the order of 10 27 N. 

4.2 Acceleration due to gravity 

Galileo was the first to make a systematic study of the motion of 
a body under the gravity of the Earth. He dropped various objects from 
the leaning tower of Pisa and made analysis of their motion under 
gravity. He came to the conclusion that "in the absence of air, all bodies 
will fall at the same rate". It is the air resistance that slows down a piece 
of paper or a parachute falling under gravity. If a heavy stone and a 
parachute are dropped where there is no air, both will fall together at 
the same rate. 

Experiments showed that the velocity of a freely falling body under 

151 



gravity increases at a constant rate, (i.e) with a constant acceleration. 
The acceleration produced in a body on account of the force of gravity 
is called acceleration due to gravity. It is denoted by g. At a given place, 
the value of g is the same for all bodies irrespective of their masses. It 
differs from place to place on the surface of the Earth. It also varies with 
altitude and depth. 

The value of g at sea-level and at a latitude of 45° is taken as the 
standard (i.e) g = 9.8 m s~ 2 

4.3 Acceleration due to gravity at the surface of the Earth 

Consider a body of mass m on the surface of 
the Earth as shown in the Fig. 4.3. Its distance 
from the centre of the Earth is R (radius of the 
Earth). 

The gravitational force experienced by the 

GMm 
body is F = 2 where M is the mass of the 

Earth. 
Fig. 4.3 Acceleration 

due to gravity From Newton's second law of motion, 

Force F = mg. 

GMm 
Equating the above two forces, 2 = rn 9 

GM 

This equation shows that g is independent of the mass of the body 

m. But, it varies with the distance from the centre of the Earth. If the 

Earth is assumed to be a sphere of radius R, the value of g on the 

GM 
surface of the Earth is given by g = -ry- 

K 

4.3.1 Mass of the Earth 

GM 
From the expression g = p2 , the mass of the Earth can be 

calculated as follows : 

qR 2 9.8 x(6.38 xlO 6 ) 2 

M =^— = pr— = 5.98 x 10 24 kg 

G 6.67XKT 11 S 




152 



4.4 Variation of acceleration due to gravity 
(i) Variation of g with altitude 

Let P be a point on the surface of the Earth and Q be a point at 
an altitude h. Let the mass of the Earth be M and radius of the Earth 
be R. Consider the Earth as a spherical shaped body. 



The acceleration due to gravity at P on the surface is 

GM 



9 



R' 



... (1) 



Let the body be placed at Q at a height h from the surface of the 
Earth. The acceleration due to gravity at Q is 

Gl 
9 h = 




(R+hr 



dividing (2) by (1) 



(2) 



9 h _ R A 



9 (R + hr 



By simplifying and expanding using 



binomial theorem, 



9k = 9 



( 

1- 

V 


2h l 


R J 



The value of acceleration due to gravity 

Fig. 4.4 Variation of g decreases with increase in height above the 

with altitude c c j, „ .-. 

surface oi the Earth. 

(ii) Variation of g with depth 

Consider the Earth to be a 
homogeneous sphere with uniform density 
of radius R and mass M. 

Let P be a point on the surface of the 
Earth and Q be a point at a depth d from 
the surface. 

The acceleration due to gravity at P on 



the surface is g 



R' 




If p be the density, then, the mass of 



the Earth is M = -n R 3 p 



Fig. 4.5 Variation of g 
with depth 



153 



g = -GnRp 



(1) 



The acceleration due to gravity at Q at a depth d from the surface 
of the Earth is 



GM 



9 d 



(R-d) 



where M d is the mass of the inner sphere of the Earth of radius (R- d). 

4 



M d = j k[R - d) 3 p 



g d = jGk(R- djp 

g d _Rd 



(2) 



dividing (2) by (1), 



R 



9 d = 9 1 



R 

The value of acceleration due to gravity decreases with increase of 
depth. 

(Hi) Variation of g with latitude (Non-sphericity of the Earth) 

The Earth is not a perfect sphere. It is 
an ellipsoid as shown in the Fig. 4.6. It is 
flattened at the poles where the latitude is 90° 
and bulged at the equator where the latitude 
is 0°. 

The radius of the Earth at equatorial 
plane R e is greater than the radius along the Fig46 Non-sphericity 




poles R by about 2 1 km. 
We know that g 



of the Earth 



•• 9 « ^2 

The value of g varies inversely as the square of radius of the 
Earth. The radius at the equator is the greatest. Hence the value of g 



154 



is minimum at the equator. The radius at poles is the least. Hence, the 
value of g is maximum at the poles. The value of g increases from the 
equator to the poles. 

(iv) Variation of g with latitude (Rotation of the Earth) 

Let us consider the Earth as a homogeneous sphere of mass M 
and radius R. The Earth rotates about an axis passing through its north 

and south poles. The Earth rotates from 
west to east in 24 hours. Its angular 
> p velocity is 7.3 x 10 5 rad s _1 . 

Consider a body of mass m on 
the surface of the Earth at P at a 
latitude 8. Let co be the angular velocity. 
^ The force (weight) F = mg acts along 
PO. It could be resolved into two 
rectangular components (i) mg cos 8 along 
PB and (ii) mg sin 8 along PA (Fig. 4.7). 

From the AOPB, it is found that 
BP = R cos 8. The particle describes a 
circle with B as centre and radius 
BP = R cos 8. 




Fig. 4.7 Rotation of 
the Earth 



The body at P experiences a centrifugal force (outward force) F„ 
due to the rotation of the Earth. 

(i.e) F c = mR(d 2 cos 8 . 

The net force along PC = mg cos 6 - mRco 2 cos 8 

The body is acted upon by two forces along PA and PC. 

The resultant of these two forces is 



F= y(mg sinQ) 2 +(mg cosd-mRo 2 cosQ) 2 
F = mg . h ■ 



since — t~ is very small, the term 



R 2 a) 4 cos 2 9 



can be neglected. 



The force, F = mg Jl 



2Ra 2 cos 2 6 



(1) 



155 



If g' is the acceleration of the body at P due to this force F, 
we have, F = mg' ... (2) 

by equating (2) and (1) 



mg' = mg ./I 



2 R co 2 c o s 2 



Rm 2 COS 2 6> A 



9 =9 | g 



Case (i) At the poles, 6 = 90° ; cos 8 = 

••• 9' = 9 
Case (ii) At the equator, 8 = 0; cos 8=1 



9 = 9 



f D 2 



g 



So, the value of acceleration due to gravity is maximum at the 
poles. 

4.5 Gravitational field 

Two masses separated by a distance exert gravitational forces on 
one another. This is called action at-a-distance. They interact even 
though they are not in contact. This interaction can also be explained 
with the field concept. A particle or a body placed at a point modifies a 
space around it which is called gravitational field. When another particle 
is brought in this field, it experiences gravitational force of attraction. 
The gravitational field is defined as the space around a mass in which it 
can exert gravitational force on other mass. 

4.5.1 Gravitational field intensity 

Gravitational field intensity or 
strength at a point is defined as the force 
experienced by a unit mass placed at 

that point. It is denoted by E. It is a 

, ... T , ., . . T , __i Fiq. 4.8 Gravitational field 

vector quantity. Its unit is N kg^ 1 . a J 

Consider a body of mass M placed 
at a point Q and another body of mass m placed at P at a distance r 
from Q. 

156 




€) 



The mass M develops a field E at P and this field exerts a force 
F = mE. 

The gravitational force of attraction between the masses m and 

GM m 
M is F = — r~ 
r 

The gravitational field intensity at P is E = — 

GM 
.-. E = — j- 
r 

Gravitational field intensity is the measure of gravitational field. 

4.5.2 Gravitational potential difference 

Gravitational potential difference between two points is defined as 
the amount of work done in moving unit mass 

from one point to another point against the , ^ . 

gravitational force of attraction. I • J > » 




Consider two points A and B separated N / n — M 

by a distance dr in the gravitational field. _, „ „ _ , A A . r , 

J & Fig. 4.9 Gravitational 

The work done in moving unit mass from potential difference 

A to B is dv = W A ^ B 

Gravitational potential difference dv = - E dr 

Here negative sign indicates that work is done against the 
gravitational field. 

4.5.3 Gravitational potential 

Gravitational potential at a point is defined as the amount of work 
done in moving unit mass from the point to infinity against the gravitational 
field. It is a scalar quantity. Its unit is Nmkg -1 . 



M 



4.5.4 Expression for gravitational potential at a point 

Consider a body of mass M at the 
point C. Let P be a point at a distance r 
from C. To calculate the gravitational 
potential at P consider two points A and 
B. The point A, where the unit mass is 
placed is at a distance x from C. 




Fig. 4.10 Gravitational potential 



157 



GM 
The gravitational field at A is E = — — 

The work done in moving the unit mass from A to B through a 
small distance dx is dw = dv = -E.dx 

Negative sign indicates that work is done against the gravitational 

field. 

GM 
dv = - — t~ dx 

x 

The work done in moving the unit mass from the point P to 

00 

infinity is d V = — I — — dx 

r 
GM 

r 
The gravitational potential is negative, since the work is done 

against the field, (i.e) the gravitational force is always attractive. 

4.5.5 Gravitational potential energy 

Consider a body of mass m placed at P at a distance r from the 
centre of the Earth. Let the mass of the Earth be M. 

When the mass m is at A at a 
t , , distance x from Q, the gravitational 




dx force of attraction on it due to mass M is 

GM m 
Fig. 4.11 Gravitational given by F = — 2 — 



potential energy 



X 



The work done in moving the mass 
m through a small distance dx from A to B along the line joining the 
two centres of masses m and M is dw = -F.dx 

Negative sign indicates that work is done against the gravitational 

field. 

GM m 
/. dw = - — j — • dx 
x 

The gravitational potential energy of a mass m at a distance rfrom 

another mass M is defined as the amount of work done in moving the 

mass m from a distance r to infinity. 

The total work done in moving the mass m from a distance r to 

158 



infinity is 



Jdw=-J 



GM m 



dx 



W = - GMm 






GMm 

*u = - 

r 

Gravitational potential energy is zero at infinity and decreases as 
the distance decreases. This is due to the fact that the gravitational 
force exerted on the body by the Earth is attractive. Hence the 
gravitational potential energy U is negative. 

4.5.6 Gravitational potential energy near the surface of the Earth 

Let the mass of the Earth be M and its radius be R. Consider a 
point A on the surface of the Earth and another point Bat a height h 
above the surface of the Earth. The work done in 



moving the mass m from A to B is U = U B 

i r 



u t 



u 



GMm 



'R+h) R 



U = GMm 
GMmh 



!R+h) 



u = 



R(R+h) 




If the body is near the surface of the Earth, p^g 4.12 Gravitational 
h is very small when compared with R. Hence {R+h) potential energy 



could be taken as R. 

GMmh 
u = 



near the surface of 
the Earth 



U 



mgh 



GM 



4.6 Inertial mass 

According to Newton's second law of motion [F = ma), the mass of 
a body can be determined by measuring the acceleration produced in it 

* Potential energy is represented by U (Upsilon). 

159 



by a constant force, (i.e) m = F/a. Intertial mass of a body is a measure 
of the ability of a body to oppose the production of acceleration in it by 
an external force. 

If a constant force acts on two masses m A and m B and produces 
accelerations a A and a B respectively, then, F = m A a A = m B a B 

m B a A 

The ratio of two masses is independent of the constant force. If the 
same force is applied on two different bodies, the inertial mass of the 
body is more in which the acceleration produced is less. 

If one of the two masses is a standard kilogram, the unknown 
mass can be determined by comparing their accelerations. 

4.7 Gravitational mass 

According to Newton's law of gravitation, the gravitational force on 
a body is proportional to its mass. We can measure the mass of a body 
by measuring the gravitational force exerted on it by a massive body like 
Earth. Gravitational mass is the mass of a body which determines the 
magnitude of gravitational pull between the body and the Earth. This is 
determined with the help of a beam balance. 

If F A and F B are the gravitational forces of attraction on the two 
bodies of masses m A and m B due to the Earth, then 

Gm A M Gm B M 

f a = —^2- and F B = —±— 

where M is mass of the Earth, R is the radius of the Earth and G is the 
gravitational constant. 

1Ll = Ll 

'"' m B F B 
If one of the two masses is a standard kilogram, the unknown 
mass can be determined by comparing the gravitational forces. 

4.8 Escape speed 

If we throw a body upwards, it reaches a certain height and then 
falls back. This is due to the gravitational attraction of the Earth. If we 
throw the body with a greater speed, it rises to a greater height. If the 

160 



body is projected with a speed of 11.2 km/s, it escapes from the Earth 
and never comes back. The escape speed is the minimum speed with 
which a body must be projected in order that it may escape from the 
gravitational pull of the planet. 

Consider a body of mass m placed on the Earth's surface. The 

GM m 

gravitational potential energy is E p = - — z — 

R 

where M is the mass of the Earth and R is its radius. 

If the body is projected up with a speed v , the kinetic energy is 

1 2 

Zk = 2 mVe 
:. the initial total energy of the body is 

1 2 GM m 

E i=2 mVe -~r - (1) 

If the body reaches a height h above the Earth's surface, the 
gravitational potential energy is 

GM m 
E D 



"p 



!R+h) 



Let the speed of the body at the height is v, then its kinetic energy is, 

1 2 
E = —mv 

Hence, the final total energy of the body at the height is 

1 2 G M m 
J 2 (R +h) 

We know that the gravitational force is a conservative force and 
hence the total mechanical energy must be conserved. 

,. E, = E f 



mvj 2 GMm mv 2 GMm 



i.e 



R 2 (R + h) 



The body will escape from the Earth's gravity at a height where 
the gravitational field ceases out. (i.e) h = oo • At the height h = oo, the 
speed v of the body is zero. 



161 



Thus ^l_ - ™HL = 

2 R 



2GM 

From the relation g = — 5- , we get GM = gR 2 

R 



Thus, the escape speed is v e = ^2gR 

The escape speed for Earth is 11.2 km/s, for the planet Mercury 
it is 4 km/s and for Jupiter it is 60 km/s. The escape speed for the 
moon is about 2.5 km/s. 

4.8.1 An interesting consequence of escape speed with the 
atmosphere of a planet 

We know that the escape speed is independent of the mass of the 
body. Thus, molecules of a gas and very massive rockets will require the 
same initial speed to escape from the Earth or any other planet or 
moon. 

The molecules of a gas move with certain average velocity, which 
depends on the nature and temperature of the gas. At moderate 
temperatures, the average velocity of oxygen, nitrogen and 
carbon-di-oxide is in the order of 0.5 km/s to 1 km/s and for lighter 
gases hydrogen and helium it is in the order of 2 to 3 km/s. It is clear 
that the lighter gases whose average velocities are in the order of the 
escape speed, will escape from the moon. The gravitational pull of the 
moon is too weak to hold these gases. The presence of lighter gases in 
the atmosphere of the Sun should not surprise us, since the gravitational 
attraction of the sun is very much stronger and the escape speed is very 
high about 620 km/s. 

4.9 Satellites 

A body moving in an orbit around a planet is called satellite. The 
moon is the natural satellite of the Earth. It moves around the Earth once 
in 27.3 days in an approximate circular orbit of radius 3.85 x 10 5 km. 
The first artificial satellite Sputnik was launched in 1956. India launched 
its first satellite Aryabhatta on April 19, 1975. 

162 



4.9.1 Orbital velocity 

Artificial satellites are made to revolve in an orbit at a height of 
few hundred kilometres. At this altitude, the friction due to air is 
negligible. The satellite is carried by a rocket to the desired height and 
released horizontally with a high velocity, so that it remains moving in 
a nearly circular orbit. 

The horizontal velocity that has to be imparted to a satellite at the 
determined height so that it makes a circular orbit around the planet is 
called orbital velocity. 

Let us assume that a satellite of mass m moves around the Earth 
in a circular orbit of radius r with uniform speed v . Let the satellite be 
at a height h from the surface of the Earth. Hence, r = R+h, where R 
is the radius of the Earth. 

The centripetal force required to keep the satellite in circular 



orbit is F = 



mv„ 



mv„ 



r R + h 
The gravitational force between the Earth and the satellite is 

GMm GMm 



1 ~ r z (R + hf 
For the stable orbital motion, 



HILL 



GMm 



R + h (R + h)' 



v„ 



GM 



IR + h 

Since the acceleration due to 

GM 



gravity on Earth's surface is g 



R* 




gR z 



Fig. 4.13 Orbital Velocity 



IR + h 

If the satellite is at a height of few hundred kilometres 
(say 200 km), [R+h] could be replaced by R. 

.". orbital velocity, v Q = ^g~R 

If the horizontal velocity (injection velocity) is not equal to the 
calculated value, then the orbit of the satellite will not be circular. If the 



163 



injection velocity is greater than the calculated value but not greater 
than the escape speed [v e =-s/2 uj» the satellite will move along an elliptical 
orbit. If the injection velocity exceeds the escape speed, the satellite will 
not revolve around the Earth and will escape into the space. If the 
injection velocity is less than the calculated value, the satellite will fall 
back to the Earth. 

4.9.2 Time period of a satellite 

Time taken by the satellite to complete one revolution round the 
Earth is called time period. 

circumference of the orbit 



T = 
to [R+h). 



Time period, T 

2nr 2n(R + h) 



orbital velocity 
where r is the radius of the orbit which is equal 



T = 2n (R+h) 
T = 2n 



R + h 
GM 



GM_ 
R + h 



(R + hf 



GM 



As GM = gR 2 , T = In 



(R + hf 



gR 2 



If the satellite orbits very close to the Earth, then h « R 

:. T = 2k J— 

4.9.3 Energy of an orbiting satellite 

A satellite revolving in a circular orbit round the Earth possesses 
both potential energy and kinetic energy. If h is the height of the satellite 
above the Earth's surface and R is the radius of the Earth, then the 
radius of the orbit of satellite is r = R+h. 

If m is the mass of the satellite, its potential energy is, 
-GMm _ -GMm 
p ~ r (R + h) 

where M is the mass of the Earth. The satellite moves with an orbital 



velocity of v Q = 



GM 
(R + h) 



164 



1 2 GMm 
Hence, its kinetic energy is, E„ = — mv E„ = — ; 

The total energy of the satellite is, E = E p + E K 

GMm 
E = ~ 2(R + h) 

The negative value of the total energy indicates that the satellite 
is bound to the Earth. 

4.9.4 Geostationary satellites 

A geo-stationary satellite is a particular type used in television and 
telephone communications. A number of communication satellites which 
appear to remain infixed positions at a specified height above the equator 
are called synchronous satellites or geo- stationary satellites. Some 
television programmes or events occuring in other countries are often 
transmitted 'live' with the help of these satellites. 

For a satellite to appear fixed at a position above a certain place 
on the Earth, its orbital period around the Earth must be exactly equal 
to the rotational period of the Earth about its axis. 

Consider a satellite of mass m moving in a circular orbit around the 
Earth at a distance rfrom the centre of the Earth. For synchronisation, its 
period of revolution around the Earth must be equal to the period of rotation 
of the Earth (ie) 1 day = 24 hr = 86400 seconds. 

The speed of the satellite in its orbit is 

Circumference of orbit 

Time period 
2nr 

v= ^r 

2 

The centripetal force is F = 



4mn 2 r 



rp2 

The gravitational force on the satellite due to the Earth is 

_ GMm 

4mn 2 r GMm . . <, _ GMT 2 
For the stable orbital motion 2 — = — 2 — l ' ~ 4 K 2 



165 



GM 
We know that, q = —5- 

.-. r 3 - 9R2T2 



4u 2 



gF^T 2 



47T 2 



The orbital radius of the geo- stationary satellite is, r = 

This orbit is called parking orbit of the satellite. 

Substituting T = 86400 s, R = 6400 km and g = 9.8 m/s 2 , the 
radius of the orbit of geo -stationary satellite is calculated as 42400 km. 

/. The height of the geo-stationary satellite above the surface of 
the Earth is h = r - R = 36000 km. 

If a satellite is parked at this height, it appears to be stationary. 
Three satellites spaced at 120° intervals each above Atlantic, Pacific and 
Indian oceans provide a worldwide communication network. 

4.9.5 Polar satellites 

The polar satellites revolve around the Earth in a north-south 
orbit passing over the poles as the Earth spins about its north - south 
axis. 

The polar satellites positioned nearly 500 to 800 km above the 
Earth travels pole to pole in 102 minutes. The polar orbit remains fixed 
in space as the Earth rotates inside the orbit. As a result, most of the 
earth's surface crosses the satellite in a polar orbit. Excellent coverage 
of the Earth is possible with this polar orbit. The polar satellites are 
used for mapping and surveying. 

4.9.6 Uses of satellites 

(i) Satellite communication 

Communication satellites are used to send radio, television and 
telephone signals over long distances. These satellites are fitted with 
devices which can receive signals from an Earth - station and transmit 
them in different directions. 

(ii) Weather monitoring 

Weather satellites are used to photograph clouds from space and 
measure the amount of heat reradiated from the Earth. With this 
information scientists can make better forecasts about weather. You 

166 



might have seen the aerial picture of our country taken by the satellites, 
which is shown daily in the news bulletin on the television and in the 
news papers. 

(Hi) Remote sensing 

Collecting of information about an object without physical contact 
with the object is known as remote sensing. Data collected by the 
remote sensing satellities can be used in agriculture, forestry, drought 
assessment, estimation of crop yields, detection of potential fishing zones, 
mapping and surveying. 

(iv) Navigation satellites 

These satellites help navigators to guide their ships or planes in 
all kinds of weather. 

4.9.7 Indian space programme 

India recognised the importance of space science and technology 
for the socio-economic development of the society soon after the launch 
of Sputnik by erstwhile USSR in 1957. The Indian space efforts started 
in 1960 with the establishment of Thumba Equatorial Rocket Launching 
Station near Thiruvananthapuram for the investigation of ionosphere. 
The foundation of space research in India was laid by Dr. Vikram 
Sarabai, father of the Indian space programme. Initially, the space 
programme was carried out by the Department of Atomic Energy. A 
separate Department of Space (DOS) was established in June 1972. 
Indian Space Research Organisation (ISRO) under DOS executes space 
programme through its establishments located at different places in 
India (Mahendragiri in Tamil Nadu, Sriharikota in Andhra Pradesh, 
Thiruvananthapuram in Kerala, Bangalore in Karnataka, Ahmedabad in 
Gujarat, etc.). India is the sixth nation in the world to have the capability 
of designing, constructing and launching a satellite in an Earth orbit. 
The main events in the history of space research in India are given below: 

Indian satellites 

1. Aryabhatta - The first Indian satellite was launched on April 19, 
1975. 

2. Bhaskara - 1 



167 



3. Rohini 

4. APPLE - It is the abbreviation of Ariane Passenger Pay Load 
Experiment. APPLE was the first Indian communication satellite put in 
geo - stationary orbit. 

5. Bhaskara - 2 

6. INSAT - 1A, IB, 1C, ID, 2A, 2B, 2C, 2D, 3A, 3B, 3C, 3D, 3E 
(Indian National Satellite). Indian National Satellite System is a joint 
venture of Department of Space, Department of Telecommunications, 
Indian Meteoro-logical Department and All India Radio and Doordarshan. 

7. SROSS - A, B, C and D (Stretched Rohini Satellite Series) 

8. IRS - 1A, IB, 1C, ID, P2, P3, P4, P5, P6 (Indian Remote Sensing 
Satellite) 

Data from IRS is used for various applications like drought 
monitoring, flood damage assessment, flood risk zone mapping, urban 
planning, mineral prospecting, forest survey etc. 

9. METSAT (Kalpana - I) - METSAT is the first exclusive 
meteorological satellite. 

10. GSAT-1, GSAT-2 (Geo-stationary Satellites) 

Indian Launch Vehicles (Rockets) 

1. SLV - 3 - This was India's first experimental Satellite Launch 
Vehicle. SLV - 3 was a 22 m long, four stage vehicle weighing 17 tonne. 
All its stages used solid propellant. 



Indian space programme is driven by the vision of Or Vikram 
Sarabhai, father of the Indian Space Programme. 

"There are some who question the relevance of space 
activities in a developing nation. To us, there is no ambiguity of 
purpose. We do not have the fantasy of competing with the 
economically advanced nations in the exploration of the moon or 
the planets or manned space- flight. But we are convinced that if 
we are to play a meaningful role nationally, and in the community 
of nations, we must be second to none in the application of 
advanced technologies to the real problems of man and society. " 




168 



2. ASLV - Augmented Satellite Launch Vehicle. It was a five stage 
solid propellant vehicle, weighing about 40 tonnes and of about 23.8 m 
long. 

3. PSLV - The Polar Satellite Launch Vehicle has four stages using 
solid and liquid propellant systems alternately. It is 44.4 m tall weighing 
about 294 tonnes. 

4. GSLV - The Geosynchronous Satellite Launch Vehicle is a 49 
m tall, three stage vehicle weighing about 414 tonnes capable of placing 
satellite of 1800 kg. 

India's first mission to moon 

ISRO has a plan to send an unmanned spacecraft to moon in the 
year 2008. The spacecraft is named as CHANDRAYAAN- 1 . This programme 
will be much useful in expanding scientific knowledge about the moon, 
upgrading India's technological capability and providing challenging 
opportunities for planetory research for the younger generation. This 
journey to moon will take 5% days. 

CHANDRAYAAN - 1 will probe the moon by orbiting it at the lunar 
orbit of altitude 100 km. This mission to moon will be carried by PSLV 
rocket. 

4.9.8 Weightlessness 

Television pictures and photographs show astronauts and objects 
floating in satellites orbiting the Earth. This apparent weightlessness is 
sometimes explained wrongly as zero-gravity condition. Then, what should 
be the reason? 

Consider the astronaut standing on the ground. He exerts a force 
(his weight) on the ground. At the same time, the ground exerts an 
equal and opposite force of reaction on the astronaut. Due to this force 
of reaction, he has a feeling of weight. 

When the astronaut is in an orbiting satellite, both the satellite 
and astronaut have the same acceleration towards the centre of the 
Earth. Hence, the astronaut does not exert any force on the floor of the 
satellite. So, the floor of the satellite also does not exert any force of 
reaction on the astronaut. As there is no reaction, the astronaut has a 
feeling of weightlessness. 

169 



4.9.9 Rockets - principle 

A rocket is a vehicle which propels itself by ejecting a 
part of its mass. Rockets are used to carry the payloads 
(satellites). We have heard of the PSLV and GSLV rockets. 
All of them are based on Newton's third law of motion. 

Consider a hollow cylindrical vessel closed on both 
ends with a small hole at one end, containing a mixture of 
combustible fuels (Fig. 4.14). If the fuel is ignited, it is 
converted into a gas under high pressure. This high pressure 
pushes the gas through the hole with an enormous force. 
This force represents the action A. Hence an opposite force, 
which is the reaction R, will act on the vessel and make it 
to move forward. 

The force (F m ) on the escaping mass of gases and 
hence the rocket is proportional to the product of the mass 

of the gases discharged per unit time — 
with which they are expelled (u) 




and the velocity 



(l.e) 



a 



dm 
~dT 



F a — (mv) 
dt 



A 
Fig. 4.14 

Principle 

of Rocket 



This force is known as momentum thrust. If the 
pressure [PJ of the escaping gases differs from the pressure 
[PJ in the region outside the rocket, there is an additional thrust called 
the velocity thrust [FJ acts. It is given by F v = A [P e - PJ where A is the 
area of the nozzle through which the gases escape. Hence, the total 
thrust on the rocket is F = F m + F v 

4.9.10 Types of fuels 

The hot gases which are produced by the combustion of a mixture of 
substances are called propellants. The mixture contains a fuel which burns 
and an oxidizer which supplies the oxygen necessary for the burning of 
the fuel. The propellants may be in the form of a solid or liquid. 



170 



4.9.11 Launching a satellite 

To place a satellite at a height of 300 km, the launching velocity 
should atleast be about 8.5 km s" 1 or 30600 kmph. If this high velocity 
is given to the rocket at the surface of the Earth, the rocket will be 
burnt due to air friction. Moreover, such high velocities cannot be 
developed by single rocket. Hence, multistage rockets are used. 

To be placed in an orbit, a satellite must be raised to the desired 
height and given the correct speed and direction by the launching rocket 
(Fig. 4.15). 

At lift off, the rocket, with a manned or unmanned satellite on top, 
is held down by clamps on the launching pad. Now the exhaust gases 
built-up an upward thrust which exceeds the rocket's weight. The clamps 
are then removed by remote control and the rocket accelerates upwards. 



o 





Satellite 



Third Stage 



Second Stage 

Combustion 
chamber 



First Stage 

Combustion 
chamber 



Satellite 

ZDP- 

Second Stage 




First Stage 



4.15 Launching a satellite 



171 



To penetrate the dense lower part of the atmosphere, initially the 
rocket rises vertically and then tilted by a guidance system. The first stage 
rocket, which may burn for about 2 minutes producing a speed of 
3 km s _1 , lifts the vehicle to a height of about 60 km and then separates 
and falls back to the Earth. 

The vehicle now goes to its orbital height, say 160 km, where it 
moves horizontally for a moment. Then the second stage of the rocket 
fires and increases the speed that is necessary for a circular orbit. By 
firing small rockets with remote control system, the satellite is separated 
from the second stage and made to revolve in its orbit. 

4.10 The Universe 

The science which deals with the study of heavenly bodies in 
respect of their motions, positions and compositions is known as 
astronomy. The Sun around which the planets revolve is a star. It is one 
of the hundred billion stars that comprise our galaxy called the Milky 
Way. A vast collection of stars held together by mutual gravitation is 
called a galaxy. The billions of such galaxies form the universe. Hence, 
the Solar system, stars and galaxies are the constituents of the universe. 

4.10.1 The Solar system 

The part of the universe in which the Sun occupies the central 
position of the system holding together all the heavenly bodies such as 
planets, moons, asteroids, comets ... etc., is called Solar system. The 
gravitational attraction of the Sun primarily governs the motion of the 
planets and other heavenly bodies around it. Mercury, Venus, Earth, 
Mars, Jupiter, Saturn, Uranus, Neptune and Pluto are the nine planets 
that revolve around the Sun. We can see the planet Venus in the early 
morning in the eastern sky or in the early evening in the western sky. 
The planet Mercury can also be seen sometimes after the sunset in the 
West or just before sunrise in the East. From the Earth, the planet Mars 
was visibly seen on 27 th August 2003. The planet Mars came closer to 
the Earth after 60,000 years from a distance of 380 x 10 6 km to a 
nearby distance of 55.7 x 10 6 km. It would appear again in the year 
2287. 

Some of the well known facts about the solar system have been 
summarised in the Table 4.1. 

172 





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173 



4.10.2 Planetary motion 

The ancient astronomers contributed a great deal by identifying 
the planets in the solar system and carefully plotting the variations in 
their positions of the sky over the periods of many years. These data 
eventually led to models and theories of the solar system. 

The first major theory, called the Geo-centric theory was developed 
by a Greek astronomer, Ptolemy. The Earth is considered to be the 
centre of the universe, around which all the planets, the moons and the 
stars revolve in various orbits. The great Indian Mathematician and 
astronomer Aryabhat of the 5th century AD stated that the Earth rotates 
about its axis. Due to lack of communication between the scientists of 
the East and those of West, his observations did not reach the 
philosophers of the West. 

Nicolaus Copernicus, a Polish astronomer proposed a new theory 
called Helio-centric theory. According to this theory, the Sun is at rest 
and all the planets move around the Sun in circular orbits. A Danish 
astronomer Tycho Brahe made very accurate observations of the motion 
of planets and a German astronomer Johannes Kepler analysed Brahe's 
observations carefully and proposed the empirical laws of planetary 
motion. 

Kepler's laws of planetary motion 
(i) The law of orbits 

Each planet moves in an elliptical orbit with the Sun at one focus. 

A is a planet revolving round . 

the Sun. The position P of the planet ^ 9 ^^^ 

where it is very close to the Sun is P /"^ Major axis\ 9 

known as perigee and the position T~ 9 ~ j 

Q of the planet where it is farthest ^^^^^^ ^^^^ 

from the Sun is known as apogee. „. „ ,„ T r ... 

r to Fig. 4.16 Law of orbits 

(ii) The law of areas 

The line joining the Sun and the planet (i.e radius vector) sweeps 
out equal areas in equal interval of times. 

The orbit of the planet around the Sun is as shown in Fig. 4.17. 
The areas A 1 and A^ are swept by the radius vector in equal times. The 
planet covers unequal distances Sj and S 2 in equal time. This is due to 

174 



the variable speed of the planet. 
When the planet is closest to the 
Sun, it covers greater distance 
in a given time. Hence, the speed 
is maximum at the closest 
position. When the planet is far 
away from the Sun, it covers 



Lov^e 




Higher 



s PeecJ 
Fig. 4.17 Law of areas 



lesser distance in the same time. Hence the speed is minimum at the 
farthest position. 

Proof for the law of areas 

Consider a planet moving from A to B. The radius vector OA 
sweeps a small angle dO at the centre in a small interval of time dt. 

From the Fig. 4.18, AB = rd 6. The small area dA swept by the 
radius is, 



dA 



1 



x r x rdQ 



Dividing by dt on both sides 

dA 

~dl 

dA 1 2 

-r co where co is 



1 2 d6 
— xr x — 

2 dt 




Fig. 4.18 Proof for the law 
of areas 



dt 2 
the angular velocity. 

The angular momentum is given by L = m/^co 

:. r z co = — 
m 

dA 1 L 

Hence, — — = — — 

dt 2 m 

Since the line of action of gravitational force passes through the 

axis, the external torque is zero. Hence, the angular momentum is 

conserved. 

dA 
•'• — — = constant. 
dt 

(i.e) the area swept by the radius vector in unit time is the same. 



(Hi) The law of periods 

The square of the period of revolution of a planet around the Sun 



175 



is directly proportional to the cube of the mean distance between the 
planet and the Sun. 

(i.e) T 2 a r 3 



—5- = constant 
r J 

Proof for the law of periods 

Let us consider a planet P of mass m moving with the velocity v 
around the Sun of mass M in a circular orbit of radius r. 

The gravitational force of attraction of the Sun on the planet is, 

GMm 



F = 

r~ 

The centripetal force is, F 
Equating the two forces 



Planet 



mv 



mu' 


GMm 


r 


r 2 


v 2 = 


GM 



.(1) 



If T be the period of revolution of the 
planet around the Sun, then 

2m 




• (2) 



Fig. 4.19 Proof for the 
law of periods 



Substituting (2) in (1) 



GM 



4K 2 r 2 



GM 



4tt' 



GM is a constant for any planet 
.-. T 2 a r 3 

4.10.3 Distance of a heavenly body in the Solar system 

The distance of a planet can be accurately measured by the radar 
echo method. In this method, the radio signals are sent towards the 
planet from a radar. These signals are reflected back from the surface 
of a planet. The reflected signals or pulses are received and detected on 

176 




Earth. The time t taken by the signal in going to the planet and coming 

back to Earth is noted. The signal travels with the velocity of the light c. 

Ct 
The distance s of the planet from the Earth is given by s 

4.10.4 Size of a planet 

It is possible to determine the size of any planet 
once we know the distance S of the planet. The image of 
every heavenly body is a disc when viewed through a 
optical telescope. The angle 8 between two extreme points 
A and B on the disc with respect to a certain point on the 
Earth is determined with the help of a telescope. The 
angle 9 is called the angular diameter of the planet. The 
linear diameter d of the planet is then given by 

d = distance x angular diameter 

d = s x Q Fig. 4.20 Size 

of a planet 

4.10.5 Surface temperatures of the planets 

The planets do not emit light of their own. They reflect the Sun's 
light that falls on them. Only a fraction of the solar radiation is absorbed 
and it heats up the surface of the planet. Then it radiates energy. We 
can determine the surface temperature T of the planet using Stefan's 
law of radiation E = a T 4 where a is the Stefan's constant and E is the 
radiant energy emitted by unit area in unit time. 

In general, the temperature of the planets decreases as we go 
away from the Sun, since the planets receive less and less solar energy 
according to inverse square law. Hence, the planets farther away from 
the Sun will be colder than those closer to it. Day temperature of 
Mercury is maximum (340°C) since it is a planet closest to the Sun and 
that of Pluto is minimum (-240°C). However Venus is an exception as 
it has very thick atmosphere of carbon-di-oxide. This acts as a blanket 
and keeps its surface hot. Thus the temperature of Venus is comparitively 
large of the order of 480°C. 

4.10.6 Mass of the planets and the Sun 

In the universe one heavenly body revolves around another massive 
heavenly body. (The Earth revolves around the Sun and the moon revolves 



177 



around the Earth). The centripetal force required by the lighter body to 
revolve around the heavier body is provided by the gravitational force of 
attraction between the two. For an orbit of given radius, the mass of the 
heavier body determines the speed with which the lighter body must 
revolve around it. Thus, if the period of revolution of the lighter body 
is known, the mass of the heavier body can be determined. For example, 
in the case of Sun - planet system, the mass of the Sun M can be 
calculated if the distance of the Sun from the Earth r, the period of 
revolution of the Earth around the Sun T and the gravitational constant 

4tt 2 r 3 

G are known using the relation M = — — 2" 

G T 

4.10.7 Atmosphere 

The ratio of the amount of solar energy reflected by the planet to 
that incident on it is known as albedo. From the knowledge of albedo, 
we get information about the existence of atmosphere in the planets. 
The albedo of Venus is 0.85. It reflects 85% of the incident light, the 
highest among the nine planets. It is supposed to be covered with thick 
layer of atmosphere. The planets Earth, Jupiter, Saturn, Uranus and 
Neptune have high albedoes, which indicate that they possess 
atmosphere. The planet Mercury and the moon reflect only 6% of the 
sunlight. It indicates that they have no atmosphere, which is also 
confirmed by recent space probes. 

There are two factors which determine whether the planets have 
atmosphere or not. They are (i) acceleration due to gravity on its surface 
and (ii) the surface temperature of the planet. 

The value of g for moon is very small (Vith of the Earth). 
Consequently the escape speed for moon is very small. As the average 
velocity of the atmospheric air molecules at the surface temperature of 
the moon is greater than the escape speed, the air molecules escape. 

Mercury has a larger value of g than moon. Yet there is no 
atmosphere on it. It is because, Mercury is very close to the Sun and 
hence its temperature is high. So the mean velocity of the gas molecules 
is very high. Hence the molecules overcome the gravitational attraction 
and escape. 



178 



4.10.8 Conditions for life on any planet 

The following conditions must hold for plant life and animal life to 
exist on any planet. 

(i) The planet must have a suitable living temperature range. 

(ii) The planet must have a sufficient and right kind of atmosphere. 

(iii) The planet must have considerable amount of water on its 
surface. 

4.10.9 Other objects in the Solar system 

(i) Asteroids 

Asteroids are small heavenly bodies which orbit round the Sun 
between the orbits of Mars and Jupiter. They are the pieces of much 
larger planet which broke up due to the gravitational effect of Jupiter. 
About 1600 asteroids are revolving around the Sun. The largest among 
them has a diameter of about 700 km is called Ceres. It circles the Sun 
once in every 4Vi years. 

(ii) Comets 

A comet consists of a small mass of rock-like material surrounded 
by large masses of substances such as water, ammonia and methane. 
These substances are easily vapourised. Comets move round the Sun in 
highly elliptical orbits and most of the time they keep far away from the 
Sun. As the comet approaches the Sun, it is heated by the Sun's radiant 
energy and vapourises and forms a head of about 10000 km in diameter. 
The comet also develops a tail pointing away from the Sun. Some comets 
are seen at a fixed regular intervals of time. Halley's comet is a periodic 
comet which made its appearance in 1910 and in 1986. It would appear 
again in 2062. 

(iii) Meteors and Meteorites 

The comets break into pieces as they approach very close to the 
Sun. When Earth's orbit cross the orbit of comet, these broken pieces 
fall on the Earth. Most of the pieces are burnt up by the heat generated 
due to friction in the Earth's atmosphere. They are called meteors 
(shooting stars). We can see these meteors in the sky on a clear moonless 
night. 

179 



Some bigger size meteors may survive the heat produced by friction 
and may not be completely burnt. These blazing objects which manage 
to reach the Earth are called meteorites. 

The formation of craters on the surface of the moon, Mercury and 
Mars is due to the fact that they have been bombarded by large number 
of meteorites. 

4.10.10 Stars 

A star is a huge, more or less spherical mass of glowing gas 
emitting large amount of radiant energy. Billions of stars form a galaxy. 
There are three types of stars. They are (i) double and multiple stars (ii) 
intrinsically variable stars and (iii) Novae and super novae. 

In a galaxy, there are only a few single stars like the Sun. Majority 
of the stars are either double stars (binaries) or multiple stars. The 
binary stars are pairs of stars moving round their common centre of 
gravity in stable equilibrium. An intrinsically variable star shows variation 
in its apparent brightness. Some stars suddenly attain extremely large 
brightness, that they may be seen even during daytime and then they 
slowly fade away. Such stars are called novae. Supernovae is a large 
novae. 

The night stars in the sky have been given names such as Sirius 
(Vyadha), Canopas (Agasti), Spica (Chitra), Arcturus (Swathi), Polaris 
(Dhruva) ... etc. After the Sun, the star Alpha Centauri is nearest to 
Earth. 

Sun 

The Sun is extremely hot and self-luminous body. It is made of 
hydrogeneous matter. It is the star nearest to the Earth. Its mass is 
about 1.989 x 10 30 kg. Its radius is about 6.95 x 10 8 m. Its distance 
from the Earth is 1.496 x 10 11 m. This is known as astronomical 
unit (AU). Light of the sun takes 8 minutes 20 seconds to reach the 
Earth. The gravitational force of attraction on the surface of the Sun is 
about 28 times that on the surface of the Earth. 

Sun rotates about its axis from East to West. The period of 
revolution is 34 days at the pole and 25 days at the equator. The density 
of material is one fourth that of the Earth. The inner part of the Sun 

180 



is a bright disc of temperature 14 x 10 6 K known as photosphere. The 
outer most layer of the Sun of temperature 6000 K is called chromosphere. 

4.10.11 Constellations 

Most of the stars appear to be grouped together forming interesting 
patterns in the sky. The configurations or groups of star formed in the 
patterns of animals and human beings are called constellations. There 
are 88 constellations into which the whole sky has been divided. 

If we look towards the northern sky on a clear moonless night 
during the months of July and August, a group of seven bright stars 
resembling a bear, the four stars forming a quadrangle form the body, 
the remaining three stars make the tail and some other faint stars form 
the paws and head of the bear. This constellation is called Ursa Major 
or Saptarishi or Great Bear. The constellation Orion resembles the figure 
of a hunter and Taurus (Vrishabha) resembles the shape of a bull. 

4.10.12 Galaxy 

A large band of stars, gas and dust particles held together by 
gravitational forces is called a galaxy. Galaxies are really complex in 
nature consisting of billions of stars. Some galaxies emit a comparatively 
small amount of radio radiations compared to the total radiations emitted. 
They are called normal galaxies. Our galaxy Milky Way is a normal 
galaxy spiral in shape. 

The nearest galaxy to us known as Andromeda galaxy, is also a 
normal galaxy. It is at a distance of 2 x 10 6 light years. (The distance 
travelled by the light in one year [9.467 x 10 12 km] is called light year). 
Some galaxies are found to emit millions of times more radio waves 
compared to normal galaxies. They are called radio galaxies. 

4.10.13 Milky Way galaxy 

Milky Way looks like a stream of milk across the sky. Some of the 
important features are given below. 

(i) Shape and size 

Milky Way is thick at the centre and thin at the edges. The 
diameter of the disc is 10 5 light years. The thickness of the Milky Way 
varies from 5000 light years at the centre to 1000 light years at the 

181 




Galactic centre 



27,000 
Light years 

Fig. 4.21 Milky Way galaxy 



position of the Sun and to k 10 5 Light years- 

500 light years at the edges. 
The Sun is at a distance of 
about 27000 light years 
from the galactic centre. 

(ii) Interstellar matter 

The interstellar space 
in the Milky Way is filled 
with dust and gases called 
inter stellar matter. It is found that about 90% of the matter is in the 
form of hydrogen. 

(Hi) Clusters 

Groups of stars held by mutual gravitational force in the galaxy 
are called star clusters. A star cluster moves as a whole in the galaxy. 
A group of 100 to 1000 stars is called galactic cluster. A group of about 
10000 stars is called globular cluster. 

(iv) Rotation 

The galaxy is rotating about an axis passing through its centre. All 
the stars in the Milky Way revolve around the centre and complete one 
revolution in about 300 million years. The Sun, one of the many stars 
revolves around the centre with a velocity of 250 km/s and its period 
of revolution is about 220 million years. 

(v) Mass 

The mass of the Milky Way is estimated to be 3 x 10 41 kg. 

4.10.14 Origin of the Universe 

The following three theories have been proposed to explain the 
origin of the Universe. 

(i) Big Bang theory 

According to the big bang theory all matter in the universe was 
concentrated as a single extremely dense and hot fire ball. An explosion 
occured about 20 billion years ago and the matter was broken into 
pieces, thrown off in all directions in the form of galaxies. Due to 



182 



continuous movement more and more galaxies will go beyond the 
boundary and will be lost. Consequently, the number of galaxies per 
unit volume will go on decreasing and ultimately we will have an empty 
universe. 

(ii) Pulsating theory 

Some astronomers believe that if the total mass of the universe is 
more than a certain value, the expansion of the galaxies would be 
stopped by the gravitational pull. Then the universe may again contract. 
After it has contracted to a certain critical size, an explosion again 
occurs. The expansion and contraction repeat after every eight billion 
years. Thus we may have alternate expansion and contraction giving 
rise to a pulsating universe. 

(Hi) Steady state theory 

According to this theory, new galaxies are continuously created 
out of empty space to fill up the gap caused by the galaxies which 
escape from the observable part of the universe. This theory, therefore 
suggests that the universe has always appeared as it does today and the 
rate of expansion has been the same in the past and will remain the 
same in future. So a steady state has been achieved so that the total 
number of galaxies in the universe remains constant. 



183 



Solved Problems 

4. 1 Calculate the force of attraction between two bodies, each of mass 
200 kg and 2 m apart on the surface of the Earth. Will the force 
of attraction be different, if the same bodies are placed on the 
moon, keeping the separation same? 

Data : m = m = 200 kg ; r = 2 m ; G = 6.67 x 10 n N m 2 kg 2 ; 
F = ? 

Gm 1 m 2 6.67 xl0' n x 200x200 
Solution : F = s ~^w? 

Force of attraction, F = 6.67 x 10 7 N 

The force of attraction on the moon will remain same, since G is the 
universal constant and the masses do not change. 

4.2 The acceleration due to gravity at the moon's surface is 1.67 ms .If 
the radius of the moon is 1.74 x 10 m, calculate the mass of the 



moon. 




Data : g = 


1.67 ms 2 ; R = 1.74 x 10 6 m ; 


G = 


6.67 x 10~ u N m 2 kg 2 ; M = ? 


Solution : 


gR 2 1.67 x (1.74 xlO 6 ) 2 
M = = 

G 6.67xl0" u 



M = 7.58 x 10 22 kg 

4.3 Calculate the height above the Earth's surface at which the value 
of acceleration due to gravity reduces to half its value on the 
Earth's surface. Assume the Earth to be a sphere of radius 
6400 km. 

9 s 

Data : h = ?; g = 77; R = 6400 x 10 m 

n, 

Solution : 



9h_ R 2 


( R 


g (R + h) 2 


{R + h 


9 J R f 




2g [R + h) 




R 1 

R + h 42 





184 



h = (^2-1) R = (1.414 - 1) 6400 x 10 3 

h = 2649.6 x 10 3 m 

At a height of 2649. 6 km from the Earth's surface, the acceleration 
due to gravity will be half of its value at the Earth's surface. 

4.4 Determine the escape speed of a body on the moon. Given : radius 
of the moon is 1.74 x 10 m and mass of the moon is 



7.36 


x 10 22 


: kg. 














Date 


i : G 


= 6.67 


x 10 11 


N m kg 2 ; 


R = 1. 


74 


x 10 6 m ; 




M 
tion : 


= 7.36 


x 10 22 


kg; 


v = ? 

e 








Solu 


<2GM 


[ 2 


x6.67 


x 10 JJ 


x7 


.36 xlO 22 


\ 




J.74x 


10' 


3 



v = 2.375 km s 1 

e 

4.5 The mass of the Earth is 81 times that of the moon and the 
distance from the centre of the Earth to that of the moon is about 
4 x 10 km. Calculate the distance from the centre of the Earth 
where the resultant gravitational force becomes zero when a 
spacecraft is launched from the Earth to the moon. 



Solution : ^p~ 




x M E 

Let the mass of the spacecraft he m. The gravitational force on the 
spacecraft at S due to the Earth is opposite in direction to that of the 
moon. Suppose the spacecraft S is at a distance xfrom the centre 
of the Earth and at a distance of (4 x 10 - x) from the moon. 

GM E m _ GM m m 



x 2 (4xl0 5 -xf 



= 81 = 



,5 ,,|2 



M m Wi (4xl0 5 -x) 

.: x = 3.6 x 10 5 km. 
The resultant gravitational force is zero at a distance of 
3.6 x 10 km from the centre of the Earth. The resultant force on S 
due to the Earth acts towards the Earth until 3.6 x 10 km is reached. 
Then it acts towards the moon. 

185 



4.6 A stone of mass 12 kg falls on the Earth's surface. If the mass of 

24 

the Earth is about 6 x 10 kg and acceleration due to gravity is 

-2 

9.8 m s , calculate the acceleration produced on the Earth by the 
stone. 

Data : m = 12 kg; M = 6 x 10 24 kg; 

a = a = 9.8 ms ; a = ? 
^ s E 

Solution : Let F be the gravitational force between the stone and 

the Earth. 

The acceleration of the stone (g) a = F/m 

The acceleration of the Earth, a = F/M 



O-E 


- m - 1J - 9 ,n- M 


a s 


M 6xl0 24 2 X W 


a 

E 


= 2 x 10~ 24 x 9.8 


a 

E 


= 19.6 x 10' 24 m s' 2 



4.7 The maximum height upto which astronaut can jump on the Earth 
is 0.75 m. With the same effort, to what height can he jump on 
the moon? The mean density of the moon is (2/3) that of the 
Earth and the radius of the moon is (1/4) that of the Earth. 

2 1 

Data : p = - p ■ R = -R; 

■m 3 F E m 4 E 

h = 0.75 m ; h = ? 

E m 

Solution : The astronaut of mass m jumps a height h on the Earth 
and a height h on the moon. If he gives himself the same kinetic 
energy on the Earth and on the moon, the potential energy gained 
at h and h will be the same. 

E m 

mgh = constant 



(1) 



mq h = 

in m 


mq h 

^E E 


K _9e 

h E 9m 




?.nrth n = 


GM E _ 



For the Earth, g = %- = — n G R p 



4 



186 



For the moon, q = 
9e__ r e Pe 

"9m R m Pm 



GM n 
R™ 



kGR p 



... (2) 



Equating (1) and (2) 

R E Pe 
h = D x h 

m K m Pm E 

R E „ P 



h = 



l 



R, 



4 ° 3 
h = 4.5 m 



^x 0.75 

Pe 



4.8 Three point masses, each of mass m, are placed at the vertices of 
an equilateral triangle of side a. What is the gravitational field and 
potential due to the three masses at the centroid of the triangle. 

Solution : 



The distance of each mass from the centroid 
is OA = OB = OC 



From the A ODC. cos 30° = 



a/2 
OC 



OC 



a/2 



a/ 



; 30° 743 



Similarly. OB = yr^and OA= Jk 



GM 




(i) The gravitational field E = j" 



3GM 



.'. Field at O due to A is, E = — 5— (towards A) 

3GM 
Field at O due to B is. E = — ^—(towards B) 

3GM 

Field at O due to C is. E = — ^—(towards C) 

c OL 



a/ 2 



187 



The resultant field due to E and E is 

~ B C 

E = ^E 2 + E 2 + 2EE cos 120° 

R B C B C 



E =^E 2 + E 2 -E 2 = E [;■ E = El 

R B B B B ' B C 

3GM 
The resultant field E = — ^~~ ac ^ s along OD. 

Since E along OA and E along OD are equal and opposite, the net 
gravitational field is zero at the centroid. 

GM 

(ii) The gravitational potential is, v = - 

r 

Net potential at 'O' is 

GM GM GM rzfGM GM GM) q M 



nr um um um 
j3\— + — + —\ =-3^3 



a/S a/ 43 a/ S '{.a a a ) a 

4.9 A geo-stationary satellite is orbiting the Earth at a height of 6R 
above the surface of the Earth. Here R is the radius of the Earth. 
What is the time period of another satellite at a height of 2.5R 
from the surface of the Earth? 

Data : The height of the geo-stationary satellite from the Earth's 
surface, h = 6R 

The height of another satellite from the Earth's surface, 
h = 2.5R 



Solution : The time period of a satellite is T = 2k 
T a (R+hf 



(R + h) 3 
GM 



For geostationary satellite, T a VfR + 6RT 



T a ^/(7R) 3 ... (1) 



For another satellite, T 2 a V(R + 2.5Rf 



T 2 a ^(3.5R) 3 ...(2) 



T 2 _ \(3.5R) 3 



Dividing (2) by (1) ^ ~ ^ (?R)3 = 2 ^ 

T, _ 24 

T 2 = 2^2 2^2 

T = 8 hours 29 minutes [;■ T = 24 hours) 

2 ' 1 

188 



Self evaluation 

(The questions and problems given in this self evaluation are only samples. 
In the same way any question and problem could be framed from the text 
matter. Students must be prepared to answer any question and problem 
from the text matter, not only from the self evaluation.) 

4.1 If the distance between two masses is doubled, the gravitational 
attraction between them 

(a) is reduced to half (b) is reduced to a quarter 

(c) is doubled (d) becomes four times 

4.2 The acceleration due to gravity at a height (l/20)th the radius of 
the Earth above the Earth's surface is 9 ms .Its value at a point at 
an equal distance below the surface of the Earth is 

(a) (b)9m s 2 

(c) 9.8 ms 2 (d) 9.5 ms 2 

4.3 The weight of a body at Earth's surface is W. At a depth halfway 
to the centre of the Earth, it will be 

(a) W (b) W/2 

(c) W/4 (d) W/8 

4.4 Force due to gravity is least at a latitude of 
(a) 0° (b) 45° 

(c) 60° (d) 90° 

4.5 If the Earth stops rotating, the value ofg at the equator will 
(a) increase (b) decrease 

(c) remain same (d) become zero 

4.6 The escape speed on Earth is 11.2 km s~ . Its value for a planet 
having double the radius and eight times the mass of the Earth is 
(a) 11.2 km s' 1 (b) 5.6 km s' 1 

(c) 22.4 km s' 1 (d) 44.8 km s' 1 

4.7 If r represents the radius of orbit of satellite of mass m moving 
around a planet of mass M. The velocity of the satellite is given by 



, , 2 GM 

(a) v = 

r 


fl,l GM 

(b) v = 

r 


. , 2 GMm 
(c)v= r 


(d)v = — 



189 



4.8 If the Earth is at one fourth of its present distance from the Sun, the 
duration of the year will be 

(a) one fourth of the present year 

(b) half the present year 

(c) one - eighth the present year 

(d) one - sixth the present year 

4.9 Which of the following objects do not belong to the solar system? 
(a) Comets (b) Nebulae 

(c) Asteroids (d) Planets 

4.10 According to Kepler's law, the radius vector sweeps out equal areas 
in equal intervals of time. The law is a consequence of the 
conservation of 

(a) angular momentum (b) linear momentum 

(c) energy (d) all the above 

4.11 Why is the gravitational force of attraction between the two bodies 
of ordinary masses not noticeable in everyday life? 

4.12 State the universal law of gravitation. 

4.13 Define gravitational constant. Give its value, unit and dimensional 
formula. 

4.14 The acceleration due to gravity varies with (i) altitude and (ii) depth. 
Prove. 

4.15 Discuss the variation of g with latitude due to the rotation of the 
Earth. 

4.16 The acceleration due to gravity is minimum at equator and maximum 
at poles. Give the reason. 

4.17 What are the factors affecting the g' value? 

4.18 Why a man can jump higher on the moon than on the Earth? 

4.19 Define gravitational field intensity. 

4.20 Define gravitational potential. 

4.21 Define gravitational potential energy. Deduce an expression for it 
for a mass in the gravitational field of the Earth. 

4.22 Obtain an expression for the gravitational potential at a point. 

4.23 Differentiate between inertial mass and gravitational mass. 

190 



4.24 The moon has no atmosphere. Why? 

4.25 What is escape speed? Obtain an expression for it. 

4.26 What is orbital velocity? Obtain an expression for it. 

4.27 What will happen to the orbiting satellite, fits velocity varies? 

4.28 What are the called geo- stationary satellites? 

4.29 Show that the orbital radius of a geo- stationary satellite is 
36000 km. 

4.30 Why do the astronauts feel weightlessness inside the orbiting 
spacecraft? 

4.31 Deduce the law of periods from the law of gravitation. 

4.32 State and prove the law of areas based on conservation of angular 
momentum. 

4.33 State Helio-Centric theory. 

4.34 State Geocentric theory. 

4.35 What is solar system? 

4.36 State Kepler's laws of planetary motion. 

4.37 What is albedo? 

4.38 What are asteroids? 

4.39 What are constellations? 

4.40 Write a note on Milky Way. 
Problems 

4.41 Two spheres of masses 10 kg and 20 kg are 5 m apart. Calculate 
the force of attraction between the masses. 

4.42 What will be the acceleration due to gravity on the surface of the 

moon, if its radius is — th the radius of the Earth and its mass is 

1 4 _ 2 

— th the mass of the Earth? (Take g as 9.8 ms ) 

4.43 The acceleration due to gravity at the surface of the moon is 
1.67 ms . The mass of the Earth is about 81 times more massive 
than the moon. What is the ratio of the radius of the Earth to that of 
the moon? 

4.44 If the diameter of the Earth becomes two times its present value 
and its mass remains unchanged, then how would the weight of an 
object on the surface of the Earth be affected? 

191 



4.45 Assuming the Earth to be a sphere of uniform density, how much 
would a body weigh one fourth down to the centre of the Earth, if it 
weighed 250 N on the surface? 

4.46 What is the value of acceleration due to gravity at an altitude of 
500 km? The radius of the Earth is 6400 km. 

4.47 What is the acceleration due to gravity at a distance from the centre 
of the Earth equal to the diameter of the Earth? 

4.48 What should be the angular velocity of the Earth, so that bodies 
lying on equator may appear weightless? How many times this 
angular velocity is faster than the present angular velocity? 
(Given ; g = 9.8 ms 2 ; R = 6400 km) 

4.49 Calculate the speed with which a body has to be projected vertically 
from the Earth's surface, so that it escapes the Earth's gravitational 
influence. (R = 6.4 x 10 3 km ; g = 9.8 m s~ 2 ) 

4.50 Jupiter has a mass 318 times that of the Earth and its radius is 
11.2 times the radius of the Earth. Calculate the escape speed of a 
body from Jupiter's surface. (Given : escape speed on Earth is 11.2 
km/s) 

4.51 A satellite is revolving in circular orbit at a height of 1 000 km from 
the surface of the Earth. Calculate the orbital velocity and time of 
revolution. The radius of the Earth is 6400 km and the mass of the 
Earth is 6x 10 24 kg. 

4.52 An artificial satellite revolves around the Earth at a distance of 
3400 km. Calculate its orbital velocity and period of revolution. 
Radius of the Earth = 6400 km; g = 9.8 m s 2 . 

4.53 A satellite of 600 kg orbits the Earth at a height of 500 kmfrom its 
surface. Calculate its (i) kinetic energy (ii) potential energy and 
(Hi) total energy ( M = 6 x 10 24 kg ; R = 6.4 x 10 6 m) 

4.54 A satellite revolves in an orbit close to the surface of a planet of 
density 6300 kg m . Calculate the time period of the satellite. Take 
the radius of the planet as 6400 km. 

4.55 A spaceship is launched into a circular orbit close to the Earth's 
surface. What additional velocity has to be imparted to the spaceship 
in the orbit to overcome the gravitational pull. 
(R = 6400 km, g= 9.8 m s~ 2 ). 

192 



Answers 

4.1 (b) 4.2 (d) 4.3 (b) 

4.4 (a) 4.5 (a) 4.6 (c) 

4.7 (a) 4.8 (c) 4.9 (b) 

4.10 (a) 

4.41 53.36x 10 n N 4.42 1.96ms 2 

4.43 3.71 4.44 W/4 

4.45 187.5 N 4.46 8.27 ms 2 

4.47 2.45 ms 2 4.48 1.25 x 10 3 rad s' 1 ; 17 

4.49 11.2 km s' 1 4.50 59.67 km s' 1 ; 

4.51 7.35 kms~ ; 1 hour 45 minutes 19 seconds 

4.52 6.4 km s' 1 ; 9614 seconds 

4.53 1.74 x 10 10 J;-3.48x 10 10 J; -1.74 x 10 10 J 

4.54 4734 seconds 4.55 3.28 km s' 1 



193 



5. Mechanics of Solids and Fluids 

Matter is a substance, which has certain mass and occupies 
some volume. Matter exists in three states namely solid, liquid and 
gas. A fourth state of matter consisting of ionised matter of bare nuclei 
is called plasma. However in our forth coming discussions, we restrict 
ourselves to the first three states of matter. Each state of matter has 
some distinct properties. For example a solid has both volume and 
shape. It has elastic properties. A gas has the volume of the closed 
container in which it is kept. A liquid has a fixed volume at a given 
temperature, but no shape. These distinct properties are due to two 
factors: (i) interatomic or intermolecular forces (ii) the agitation or 
random motion of molecules due to temperature. 

In solids, the atoms and molecules are free to vibrate about their 
mean positions. If this vibration increases sufficiently, molecules will 
shake apart and start vibrating in random directions. At this stage, the 
shape of the material is no longer fixed, but takes the shape of its 
container. This is liquid state. Due to increase in their energy, if the 
molecules vibrate at even greater rates, they may break away from one 
another and assume gaseous state. Water is the best example for this 
changing of states. Ice is the solid form of water. With increase in 
temperature, ice melts into water due to increase in molecular vibration. 
If water is heated, a stage is reached where continued molecular 
vibration results in a separation among the water molecules and 
therefore steam is produced. Further continued heating causes the 
molecules to break into atoms. 



5. 1 Intermolecular or interatomic 
forces 

Consider two isolated hydrogen 
atoms moving towards each other as 
shown in Fig. 5.1. 

As they approach each other, 
the following interactions are 
observed. 




Fig. 5.1 Electrical origin of 
interatomic forces 



207 



(i) Attractive force A between the nucleus of one atom and electron 
of the other. This attractive force tends to decrease the potential energy 
of the atomic system. 

(ii) Repulsive force R between the nucleus of one atom and the 
nucleus of the other atom and electron of one atom with the electron 
of the other atom. These repulsive forces always tend to increase the 
energy of the atomic system. 

There is a universal tendency of all systems to acquire a state of 
minimum potential energy. This stage of minimum potential energy 
corresponds to maximum stability. 

If the net effect of the forces of attraction and repulsion leads to 
decrease in the energy of the system, the two atoms come closer to 
each other and form a covalent bond by sharing of electrons. On the 
other hand, if the repulsive forces are more and there is increase in the 
energy of the system, the atoms will repel each other and do not form 
a bond. 

The variation of potential energy with interatomic distance between 
the atoms is shown in Fig. 5.2. 



.2 O 




Interatomic distance between hydrogen atoms 



Fig. 5.2. Variation of potential energy with interatomic distance 



208 



It is evident from the graph that as the atoms come closer i.e. 
when the interatomic distance between them decreases, a stage is 
reached when the potential energy of the system decreases. When the 
two hydrogen atoms are sufficiently closer, sharing of electrons takes 
place between them and the potential energy is minimum. This results 
in the formation of covalent bond and the interatomic distance is r . 

In solids the interatomic distance is r o and in the case of liquids 
it is greater than r . For gases, it is much greater than r . 

The forces acting between the atoms due to electrostatic interaction 
between the charges of the atoms are called interatomic forces. Thus, 
interatomic forces are electrical in nature. The interatomic forces are 
active if the distance between the two atoms is of the order of atomic 
size a 10' 10 m. In the case of molecules, the range of the force is of the 
order of 10~ 9 m. 

5.2 Elasticity 

When an external force is applied on a body, which is not free 
to move, there will be a relative displacement of the particles. Due to 
the property of elasticity, the particles tend to regain their original 
position. The external forces may produce change in length, volume 
and shape of the body. This external force which produces these changes 
in the body is called deforming force. A body which experiences such 
a force is called deformed body. When the deforming force is removed, 
the body regains its original state due to the force developed within the 
body. This force is called restoring force. The property of a material to 
regain its original state when the deforming force is removed is called 
elasticity. The bodies which possess this property are called elastic 
bodies. Bodies which do not exhibit the property of elasticity are called 
plastic. The study of mechanical properties helps us to select the 
material for specific purposes. For example, springs are made of steel 
because steel is highly elastic. 

Stress and strain 

In a deformed body, restoring force is set up within the body 
which tends to bring the body back to the normal position. The 
magnitude of these restoring force depends upon the deformation 
caused. This restoring force per unit area of a deformed body is known 
as stress. 

209 



Stress 



restoring force 



area 



N rn 2 



Its dimensional formula is ML i T~ 2 . 

Due to the application of deforming force, length, volume or 
shape of a body changes. Or in other words, the body is said to be 
strained. Thus, strain produced in a body is defined as the ratio of 
change in dimension of a body to the original dimension. 

change in dimension 
. . strain - original dimension 

Strain is the ratio of two similar quantities. Therefore it has no 
unit. 



Elastic limit 

If an elastic material is stretched or compressed beyond a certain 
limit, it will not regain its original state and will remain deformed. The 
limit beyond which permanent deformation occurs is called the elastic 
limit. 

Hooke's law 

English Physicist Robert Hooke (1635 - 1703) in the year 1676 

put forward the relation between the extension produced in a wire and 

the restoring force developed in it. The law formulated on the basis of 

this study is known as Hooke's law. According to Hooke's law, within 

the elastic limit, strain produced in a body is directly proportional to the 

stress that produces it. 

(i.e) stress a strain 

■*■■■'•" 
Stress 

— — — = a constant, known as modulus of 
Strain J 

elasticity. 

Its unit is N m 2 and its dimensional formula 
is ML : T 2 . 

5.2.1 Experimental verification of Hooke's law 

A spring is suspended from a rigid support 
as shown in the Fig. 5.3. A weight hanger and a 
light pointer is attached at its lower end such 



Spring 



Slotted 
Weights 



Fig. 5.3 Experimental 

setup to verify 

Hooke's law 



210 



that the pointer can slide over a scale graduated in millimeters. The 
initial reading on the scale is noted. A slotted weight of m kg is added 
to the weight hanger and the pointer position is noted. The same 
procedure is repeated with every additional m kg weight. It will be 
observed that the extension of the spring is proportional to the weight. 
This verifies Hooke's law. 

5.2.2 Study of stress - strain relationship 

Let a wire be suspended from a rigid support. At the free end, a 
weight hanger is provided on which weights could be added to study 
the behaviour of the wire under different load conditions. The extension 
of the wire is suitably 
measured and a stress - strain 
graph is plotted as in Fig. 5.4. 

(i) In the figure the region 
OP is linear. Within a normal 
stress, strain is proportional to 
the applied stress. This is 
Hooke's law. Upto P, when the 
load is removed the wire 
regains its original length along 
PO. The point P represents the 
elastic limit, PO represents the 
elastic range of the material 
and OB is the elastic strength. 



Stress 




Fig. 5.4 Stress 



Strain 
Strain relationship 



(ii) Beyond P, the graph is not linear. In the region PQ the material 
is partly elastic and partly plastic. From Q, if we start decreasing the 
load, the graph does not come to O via P, but traces a straight line QA. 
Thus a permanent strain OA is caused in the wire. This is called 
permanent set. 

(iii) Beyond Q addition of even a very small load causes enormous 
strain. This point Q is called the yield point. The region QR is the 
plastic range. 

(iv) Beyond R, the wire loses its shape and becomes thinner and 
thinner in diameter and ultimately breaks, say at S. Therefore S is the 
breaking point. The stress corresponding to S is called breaking stress. 



211 



5.2.3 Three moduli of elasticity 

Depending upon the type of strain in the body there are three 
different types of modulus of elasticity. They are 

(i) Young's modulus 

(ii) Bulk modulus 

(iii) Rigidity modulus 

(i) Young's modulus of elasticity 

Consider a wire of length I and cross sectional area A 
stretched by a force F acting along its length. Let dl be the 
extension produced. 



Longitudinal stress = 



Force 



F 
A 



dl 



vL 

in 



Longitudinal strain = 



Area 
change in length 



m 

T 



original length 

Young's modulus of the material of the wire is defined 
as the ratio of longitudinal stress to longitudinal strain. It is 
denoted by q. 



Fig. 5.5 

Young's 

modulus of 

elasticity 



(i.e) q = 



Young's modulus 
F/A 



longitudinal stress 
longitudinal strain 

F I 



dl/l 



or 



q = 



A dl 



(ii) Bulk modulus of elasticity 

Suppose euqal forces act 

perpendicular to the six faces of a cube 

of volume V as shown in Fig. 5.6. Due 

to the action of these forces, let the 

decrease in volume be dV. 

Force F 

Now, Bulk stress = — = — 

Area A 

Bulk Strain = 

change in volume -dV 
original volume V 

(The negative sign indicates that 
volume decreases.) 











F 






a 


d 










> 


' 


yS ( 










b 










f 


'* 






h 
















y/& 




t 


71 








/ 


^ 


i 






F 









Fig. 5.6 Bulk modulus 
of elasticity 



212 



Bulk modulus of the material of the object is defined as the ratio 
bulk stress to bulk strain. 
It is denoted by k. 

Bulk stress 



(i.e) k = 



UULUUfc 


Bulk strain 








F/A 
dV 


P 
dV 


A 


or 


k = 


-PV 
~dV 


V 


V 











(Hi) 



Rigidity modulus or shear modulus 

Let us apply a force F 
tangential to the top surface of 
a block whose bottom AB Is 
fixed, as shown in Fig. 5.7. 

Under the action of this 
tangential force, the body suffers 
a slight change in shape, its 
volume remaining unchanged. 
The side AD of the block is 
sheared through an angle 9 to 
the position AD'. 



F4 




A B 

Fig. 5.7 Rigidity modulus 



If the area of the top surface is A, then shear stress = F/A. 

Shear modulus or rigidity modulus of the material of the object is 
defined as the ratio of shear stress to shear strain. It is denoted by n. 



Rigidity modulus = 
F/A 



9 



(i.e) n 

F 
~A0 

Table 5.1 lists the 
values of the three 
moduli of elasticity for 
some commonly used 
materials. 



shear stress 
shear strain 

Table 5.1 Values for the 
moduli of elasticity 



Material 


Modulus oj elasticity fx 10 n Pa) 


1 


k 


n 


Aluminium 

Copper 

Iron 

Steel 

Tungsten 


0.70 
1.1 

1.9 
2.0 
3.6 


0.70 

1.4 

1.0 

1.6 

2.0 


0.30 
0.42 
0.70 
0.84 
1.5 



213 




5.2.4 Relation between the three 
moduli of elasticity 

Suppose three stresses P, Q and 
R act perpendicular to the three faces 
ABCD, ADHE and ABFE of a cube of 
unit volume (Fig. 5.8). Each of these 
stresses will produce an extension in 
its own direction and a compression 
along the other two perpendicular 
directions. If A is the extension per unit 
stress, then the elongation along the 
direction of P will be XP. If \i is the 
contraction per unit stress, then the contraction along the direction of 
P due to the other two stresses will be y,Q and y,R. 

.". The net change in dimension along the direction of P due to all 
the stresses is e = XP - \^Q - \^R. 

Similarly the net change in dimension along the direction of Q is 
/ = XQ - aP - \y.R and the net change in dimension along the direction 
of R is g = XR - [iP - [iQ. 

Case (i) 

If only P acts and Q = R = then it is a case of longitudinal stress. 
.". Linear strain = e = XP 



Fig. 5.8 Relation between the 
three moduli of elasticity 



Young's modulus q 



linear stress 
linear strain 



P 

~XP 



(i.e) q = 



or 



X = 



l 



• (1) 



Case (ii) 

If R = O and P = - Q, then the change in dimension along P is 
e = XP - u (-P) 

(i.e) e = (X + u) P 

Angle of shear 6 = 2e* = 2 (X + u) P 
Rigidity modulus 

P P 

9 ~ 2(A + u)P 



n = 



(or) 2 (X + u) = 



1 



n 



• (2) 



* The proof for this is not given here 



214 



Case (Hi) 

If P = Q = R, the increase in volume is = e + f + g 

= 3e = 3(X - 2fi) P (since e = f = g) 
:. Bulk strain = 3{X-2\x) P 

P 



Bulk modulus k = 



From (2), 2{X + /.i) 



3 (A - 2]jl)P 
1 



or (X - 2[i) = 



n 



2X + 2fi = 



1 



n 



From (3), (X - 2/j) 



3k 



Adding (4) 


and (5), 


3X 


1 1 
n 3k 


X 


1 1 
3n 9k 


1 1 1 
.-. From (1), - = — + ;rr 
q 3n 9k 




9 3 1 

or ~ + , 
q n k 



This is the relation between the three 
moduli of elasticity. 

5.2.5 Determination of Young's modulus 
by Searle's method 

The Searle's apparatus consists of two 
rectangular steel frames A and B as shown 
in Fig. 5.9. The two frames are hinged 
together by means of a frame F. A spirit 
level L is provided such that one of its ends 
is pivoted to one of the frame B whereas the 
other end rests on top of a screw working 
through a nut in the other frame. The bottom 



1 
3k 



• (3) 



...(4) 
...(5) 



L 



F 



"^ 



i 




Fig. 5.9 Searle's 
apparatus 



215 



of the screw has a circular scale C which can move along a vertical 
scale V graduated in mm. This vertical scale and circular scale 
arrangement act as pitch scale and head scale respectively of a 
micrometer screw. 

The frames A and B are suspended from a fixed support by means 
of two wires PQ and RS respectively. The wire PQ attached to the frame 
A is the experimental wire. To keep the reference wire RS taut, a constant 
weight W is attached to the frame B. To the frame A, a weight hanger 
is attached in which slotted weights can be added. 

To begin with, the experimental wire PQ is brought to the elastic 
mood by loading and unloading the weights in the hanger in the frame 
A four or five times, in steps of 0.5 kg. Then with the dead load, the 
micrometer screw is adjusted to ensure that both the frames are at the 
same level. This is done with the help of the spirit level. The reading 
of the micrometer is noted by taking the readings of the pitch scale 
and head scale. Weights are added to the weight hanger in steps of 0.5 
kg upto 4 kg and in each case the micrometer reading is noted by 
adjusting the spirit level. The readings are again noted during unloading 
and are tabulated in Table 5.2. The mean extension dl for M kg of load 
is found out. 

Table 5.2 Extension for M kg weight 



Load in weight 
hanger kg 


Micrometer reading 


Extension 
for M kg weight 


Loading 


Unloading 


Mean 


W 

W + 0.5 
W + 1.0 
W + 1.5 
W + 2.0 
W + 2.5 
W + 3.0 
W + 3.5 
W + 4.0 











216 



If I is the original length and r the mean radius of the experimental 
wire, then Young's modulus of the material of the wire is given by 



F/A 
(i.e) q = 



F/nr 2 
dl/l 

Fl 



Kr 2 dl 



5.2.6 Applications of modulus of elasticity 

Knowledge of the modulus of elasticity of materials helps us to 
choose the correct material, in right dimensions for the right application. 
The following examples will throw light on this. 

(i) Most of us would have seen a crane used for lifting and moving 
heavy loads. The crane has a thick metallic rope. The maximum load 
that can be lifted by the rope must be specified. This maximum load 
under any circumstances should not exceed the elastic limit of the 
material of the rope. By knowing this elastic limit and the extension per 
unit length of the material, the area of cross section of the wire can be 
evaluated. From this the radius of the wire can be calculated. 

(ii) While designing a bridge, one has to keep in mind the following 
factors (1) traffic load (2) weight of bridge (3) force of winds. The bridge is 
so designed that it should neither bend too much nor break. 

5.3 Fluids 

A fluid is a substance that can flow when external force is applied 
on it. The term fluids include both liquids and 
gases. Though liquids and gases are termed 
as fluids, there are marked differences between 
them. For example, gases are compressible 
whereas liquids are nearly incompressible. We 
only use those properties of liquids and gases, 
which are linked with their ability to flow, 
while discussing the mechanics of fluids. 



5.3.1 Pressure due to a liquid column 

Let h be the height of the liquid column 
in a cylinder of cross sectional area A. If P is 
the density of the liquid, then weight of the 



Fig. 5.10 Pressure 



217 



liquid column W is given by 

W = mass of liquid column x g = Ahpg 
By definition, pressure is the force acting per unit area. 
weight of liquid column 



Pressure = 



area of cross - sec tion 
Ahpg 



A 
P =hpg 



= hpg 




5.3.2 Pascal's law 

One of the most important facts about 
fluid pressure is that a change in pressure at 
one part of the liquid will be transmitted 
without any change to other parts. This was 
put forward by Blaise Pascal (1623 - 1662), a 
French mathematician and physicist. This rule 
is known as Pascal's law. 

Pascal's law states that if the effect of 
gravity can be neglected then the pressure in a Fig. 5.11 Pascal's law in 
fluid in equilibrium is the same everywhere. tfie aDsence °f gravity 

Consider any two points A and B inside the fluid. Imagine a 
cylinder such that points A and B lie at the centre of the circular 
surfaces at the top and bottom of the cylinder (Fig. 5.11). Let the fluid 
inside this cylinder be in equilibrium under the action of forces from 
outside the fluid. These forces act everywhere perpendicular to the 
surface of the cylinder. The forces acting on the circular, top and bottom 
surfaces are perpendicular to the forces acting on the cylindrical surface. 
Therefore the forces acting on the faces at A and B are equal and 
opposite and hence add to zero. As the areas of these two faces are 
equal, we can conclude that pressure at A is equal to pressure at B. 
This is the proof of Pascal's law when the effect of gravity is not taken 
into account. 

Pascal's law and effect of gravity 

When gravity is taken into account, Pascal's law is to be modified. 
Consider a cylindrical liquid column of height h and density P in a 



218 



I 



Mf- 



-h- 



.TST-l 



vessel as shown in the Fig. 5.12. 

If the effect of gravity is neglected, then 
pressure at M will be equal to pressure at N. 
But, if force due to gravity is taken into account, 
then they are not equal. 

As the liquid column is in equilibrium, the 
forces acting on it are balanced. The vertical 
forces acting are 

(i) Force Pfi acting vertically down on the Fi g 12 Pascal's law 
top surface. and effect of gravity 

(ii) Weight mg of the liquid column acting vertically downwards, 
(iii) Force P^ at the bottom surface acting vertically upwards. 

where P 1 and P 2 are the pressures at the top and bottom faces, A is 
the area of cross section of the circular face and m is the mass of the 
cylindrical liquid column. 

At equilibrium, P^A + mg - P^ = or P^A + mg = P^ 

mg 



P > + A 



But 



m = Ahp 



Pi + 



Ahpg 



* 2 -i ■ A 

(i.e) P 2 = P 1 + hpg 

This equation proves that the pressure is the same at all points 
at the same depth. This results in another statement of Pascal's law 
which can be stated as change in pressure at any point in an enclosed 
fluid at rest is transmitted undiminished to all points in the fluid and act 
in all directions. 



5.3.3 Applications of Pascal's law 

(i) Hydraulic lift 

An important application of Pascal's law is the hydraulic lift used 
to lift heavy objects. A schematic diagram of a hydraulic lift is shown 
in the Fig. 5.13. It consists of a liquid container which has pistons 
fitted into the small and large opening cylinders. If a 1 and a 2 are the 
areas of the pistons A and B respectively, F is the force applied on A 
and W is the load on B, then 



219 



or 



W = F 



a, 



F _W 

This is the load that can be lifted 
by applying a force F on A. In the above 

is called mechanical 



w 



a Q 



equation 



a, 



advantage of the hydraulic lift. One can 
see such a lift in many automobile 
service stations. 



-B~- 



Fig. 5.13 Hydraulic lift 



(ii) Hydraulic brake 

When brakes are applied suddenly in a moving vehicle, there is 
every chance of the vehicle to skid because the wheels are not retarded 
uniformly. In order to avoid this danger of skidding when the brakes are 
applied, the brake mechanism must be such that each wheel is equally 
and simultaneously retarded. A hydraulic brake serves this purpose. It 
works on the principle of Pascal's law. 

Fig. 5.14 shows the schematic diagram of a hydraulic brake system. 
The brake system has a main cylinder filled with brake oil. The main 
cylinder is provided with a piston P which is connected to the brake 



Brake Peda 



Pipe line to 
other wheels 




Inner rim of the wheel 



Fig. 5.14 Hydraulic brake 
220 



pedal through a lever assembly. A T shaped tube is provided at the 
other end of the main cylinder. The wheel cylinder having two pistons 
Pj and P 2 is connected to the Ttube. The pistons P 1 and P 2 are connected 
to the brake shoes S 1 and S 2 respectively. 

When the brake pedal is pressed, piston P is pushed due to the 
lever assembly operation. The pressure in the main cylinder is 
transmitted to P : and P 2 . The pistons P 1 and P 2 push the brake shoes 
away, which in turn press against the inner rim of the wheel. Thus the 
motion of the wheel is arrested. The area of the pistons P 1 and P 2 is 
greater than that of P. Therefore a small force applied to the brake 
pedal produces a large thrust on the wheel rim. 

The main cylinder is connected to all the wheels of the automobile 
through pipe line for applying equal pressure to all the wheels . 

5.4 Viscosity 

Let us pour equal amounts of water and castor oil in two identical 
funnels. It is observed that water flows out of the funnel very quickly 
whereas the flow of castor oil is very slow. This is because of the 
frictional force acting within the liquid. This force offered by the adjacent 
liquid layers is known as viscous force and the phenomenon is called 
viscosity. 

Viscosity is the property of the fluid by virtue of which it opposes 
relative motion between its different layers. Both liquids and gases exhibit 
viscosity but liquids are much more viscous than gases. 

Co-efficient of viscosity 

Consider a liquid to flow 
steadily through a pipe as shown 
in the Fig. 5.15. The layers of 
the liquid which are in contact 
with the walls of the pipe have 
zero velocity. As we move towards 
the axis, the velocity of the liquid 
layer increases and the centre 



3^ 



i\ 



v-- 



'S 



Fig. 5.15 Steady flow of a liquid 



layer has the maximum velocity v. Consider any two layers P and Q 
separated by a distance dx. Let dv be the difference in velocity between 
the two layers. 



221 



The viscous force F acting tangentially between the two layers of 
the liquid is proportional to (i) area A of the layers in contact 

dv 
(ii) velocity gradient — perpendicular to the flow of liquid. 



F a A 



\] A 



dv 

dx 
dv 



dx 



where r| is the coefficient of viscosity of the liquid. 

This is known as Newton's law of viscous flow in fluids. 

If A = lm 2 and -^ = Is' 1 
dx 

then F = r\ 

The coefficient of viscosity of a liquid is numerically equal to the 
viscous force acting tangentially between two layers of liquid having unit 
area of contact and unit velocity gradient normal to the direction of flow 
of liquid. 

The unit of 77 is N s m~ 2 . Its dimensional formula is ML _J T~ J . 

5.4.1 Streamline flow 

The flow of a liquid is said to be steady, streamline or laminar if 
every particle of the liquid follows exactly the path of its preceding particle 
and has the same velocity of its preceding particle at every point. 




Fig. 5.16 Steamline flow 



Let abc be the path of flow 
of a liquid and v v v 2 and v 3 
be the velocities of the liquid 
at the points a, b and c 
respectively. During a 
streamline flow, all the particles 
arriving at 'a' will have the same 



velocity v 1 which is directed along the tangent at the point 'a'. A particle 
arriving at b will always have the same velocity v 2 . This velocity v 2 may 
or may not be equal to v 1 . Similarly all the particles arriving at the point 
c will always have the same velocity l> 3 . In other words, in the streamline 
flow of a liquid, the velocity of every particle crossing a particular point 
is the same. 



222 



The streamline flow is possible only as long as the velocity of the 
fluid does not exceed a certain value. This limiting value of velocity is 
called critical velocity. 

5.4.2 Turbulent flow 

When the velocity of a liquid exceeds the critical velocity, the 
path and velocities of the liquid become disorderly. At this stage, the 
flow loses all its orderliness and is called turbulent flow. Some examples 
of turbulent flow are : 

(i) After rising a short distance, the smooth column of smoke 
from an incense stick breaks up into irregular and random patterns. 

(ii) The flash - flood after a heavy rain. 

Critical velocity of a liquid can be defined as that velocity of liquid 
upto which the flow is streamlined and above which its flow becomes 
turbulent. 

5.4.3 Reynold's number 

Reynolds number is a pure number which determines the type of 
flow of a liquid through a pipe. It is denoted by N R . 
It is given by the formula 

R '? 

where u c is the critical velocity, p is the density, 77 is the co-efficient 
of viscosity of the liquid and D is the diameter of the pipe. 

If N R lies between and 2000, the flow of a liquid is said to be 
streamline. If the value of N R is above 3000, the flow is turbulent. If 
N R lies between 2000 and 3000, the flow is neither streamline nor 
turbulent, it may switch over from one type to another. 

Narrow tubes and highly viscous liquids tend to promote stream 
line motion while wider tubes and liquids of low viscosity lead to 
tubulence. 

5.4.4 Stoke's law (for highly viscous liquids) 

When a body falls through a highly viscous liquid, it drags the 
layer of the liquid immediately in contact with it. This results in a 
relative motion between the different layers of the liquid. As a result 
of this, the falling body experiences a viscous force F. Stoke performed 

223 



many experiments on the motion of small spherical bodies in different 
fluids and concluded that the viscous force F acting on the spherical 
body depends on 

(i) Coefficient of viscosity r| of the liquid 

(ii) Radius a of the sphere and 

(iii) Velocity v of the spherical body. 

Dimensionally it can be proved that 

F = k r\av 

Experimentally Stoke found that 

k = 6n 

F = 6k nav 
This is Stoke's law. 



5.4.5 Expression for terminal velocity 

Consider a metallic sphere of radius 'a! and 
density p to fall under gravity in a liquid of density 
G ■ The viscous force F acting on the metallic sphere 
increases as its velocity increases. A stage is reached 
when the weight W of the sphere becomes equal to 
the sum of the upward viscous force F and the upward 
thrust U due to buoyancy (Fig. 5.17). Now, there is 
no net force acting on the sphere and it moves down 
with a constant velocity v called terminal velocity. 

.-. W - F - U = O ...(1) 

Terminal velocity of a body is defined as the 

constant velocity acquired by a body while falling 

through a viscous liquid. 

From (1), W = F + U ...(2) 

According to Stoke's law, the viscous force F is 
given by F = 67ir|au. 



■■IT-: 



:Q 



:w: 



Fig. 5.17 Sphere 

falling in a 

viscous liquid 



The buoyant force U = Weight of liquid displaced by the sphere 

4 3 



The weight of the sphere 

4 „ 

W = -nar'pg 



224 



Substituting in equation (2), 



; ra pg 



6n riav + — no? a g 



or 6n nav = — no? (p-a)g 



v = 



2 a 2 (p- <j)g 



5.4.6 Experimental determination of viscosity of 
highly viscous liquids 











111 




B — 




t 






1 






1 






c — 







Fig. 5.18 
Experimental 
determination 
of viscosity of 
highly viscous 
liquid 



The coefficient of highly viscous liquid like castor 
oil can be determined by Stoke's method. The 
experimental liquid is taken in a tall, wide jar. Two 
marking B and C are marked as shown in Fig. 5.18. 
A steel ball is gently dropped in the jar. 

The marking B is made well below the free surface of the liquid 
so that by the time ball reaches B, it would have acquired terminal 
velocity v. 

When the ball crosses B, a stopwatch is switched on and the time 
taken t to reach C is noted. If the distance BC is s, then terminal 

S 

velocity v = ~ . 

The expression for terminal velocity is 



v = 



2 a 2 lp-o)g 



- 9 r. 












s 2 a 2 (p - 
t 9 ri 


a)g 


or 


r| = 


.frffc- 


o)g - 
s 



Knowing a, p and a . the value of r\ of the liquid is determined. 

Application of Stoke's law 

Falling of rain drops: When the water drops are small in size, 
their terminal velocities are small. Therefore they remain suspended in 
air in the form of clouds. But as the drops combine and grow in size, 
their terminal velocities increases because v a a 2 . Hence they start 
falling as rain. 



225 



5.4.7 Poiseuille's equation 

Poiseuille investigated the steady flow of a liquid through a capillary 
tube. He derived an expression for the volume of the liquid flowing per 
second through the tube. 

Consider a liquid of co-efficient of viscosity r| flowing, steadily 
through a horizontal capillary tube of length I and radius r. If P is the 
pressure difference across the ends of the tube, then the volume V of 
the liquid flowing per second through the tube depends on t|, r and 

( p 
the pressure gradient y 

(i.e) V a r\ x r« (y 

v = M x ry (yj ...(i) 

where k is a constant of proportionality. Rewriting equation (1) in 
terms of dimensions, 

"ml _1 t 2 

[VT 1 ] = [ML 1 T !] x [LP 

Equating the powers of L, M and T on both sides we get 
x = -1, y = 4 and z = 1 

Substituting in equation (1), 

V = k r[ l r 4 (y 



V 



kPr 4 



ijl 



Experimentally k was found to be equal to „ 



xPr 4 



v 



8tjl 

This is known as Poiseuille's equation. 

226 



5.4.8 Determination of coefficient of viscosity of water by 
Poiseuille's flow method 

A capillary tube of very fine bore 
is connected by means of a rubber tube 
to a burette kept vertically. The capillary 
tube is kept horizontal as shown in 
Fig. 5. 19. The burette is filled with water 
and the pinch - stopper is removed. 
The time taken for water level to fall 
from A to B is noted. If V is the volume 
between the two levels A and B, then 
volume of liquid flowing per second is 

V 

— . If I and r are the length and radius 

of the capillary tube respectively, then 







A 


u 



Fig. 5.19 Determination of 

coefficient of 

viscosity by Poiseuille's flow 



V 
t 



kPt* 
8t]I 



• (1) 



If p is the density of the liquid 

then the initial pressure difference 

between the ends of the tube is P 1 = ftjpgf and the final pressure difference 

P 2 = h 2 pg. Therefore the average pressure difference during the flow of 

water is P where 

» _ P l +P 2 



2 pa = hpg 



.. h = hj+h.2 



Substituting in equation (1), we get 



V 
t 



nhpgr 



or 



nhpgr t 



8lr] ~ '' 81V 

5.4.9 Viscosity - Practical applications 

The importance of viscosity can be understood from the following 
examples. 

(i) The knowledge of coefficient of viscosity of organic liquids is 
used to determine their molecular weights. 



227 



(ii) The knowledge of coefficient of viscosity and its variation with 
temperature helps us to choose a suitable lubricant for specific 
machines. In light machinery thin oils (example, lubricant oil used in 
clocks) with low viscosity is used. In heavy machinery, highly viscous 
oils (example, grease) are used. 

5.5 Surface tension 

Intermolecular forces 

The force between two molecules of a substance is called 
intermolecular force. This intermolecular force is basically electric in 
nature. When the distance between two molecules is greater, the 
distribution of charges is such that the mean distance between opposite 
charges in the molecule is slightly less than the distance between their 
like charges. So a force of attraction exists. When the intermolecular 
distance is less, there is overlapping of the electron clouds of the 
molecules resulting in a strong repulsive force. 

The intermolecular forces are of two types. They are (i) cohesive 
force and (ii) adhesive force. 

Cohesive force 

Cohesive force is the force of attraction between the molecules of 
the same substance. This cohesive force is very strong in solids, weak 
in liquids and extremely weak in gases. 

Adhesive force 

Adhesive force is the force of attraction between the moelcules of 
two different substances. For example due to the adhesive force, ink 
sticks to paper while writing. Fevicol, gum etc exhibit strong adhesive 
property. 

Water wets glass because the cohesive force between water 
molecules is less than the adhesive force between water and glass 
molecules. Whereas, mercury does not wet glass because the cohesive 
force between mercury molecules is greater than the adhesive force 
between mercury and glass molecules. 

Molecular range and sphere of influence 

Molecular range is the maximum distance upto which a molecule 
can exert force of attraction on another molecule. It is of the order of 
10~ 9 m for solids and liquids. 

228 



Sphere of influence is a sphere drawn around a particular molecule 
as centre and molecular range as radius. The central molecule exerts a 
force of attraction on all the molecules lying within the sphere of 
influence. 




5.5.1 Surface tension of a liquid 

Surface tension is the property of the free surface 
of a liquid at rest to behave like a stretched membrane 
in order to acquire minimum surface area. 

Imagine a line AB in the free surface of a liquid 
at rest (Fig. 5.20). The force of surface tension is 
measured as the force acting per unit length on either 
side of this imaginary line AB. The force is 
perpendicular to the line and tangential to the liquid 
surface. If F is the force acting on the length I of the 
line AB, then surface tension is given by 

*-f 

Surface tension is defined as the force per unit length acting 
perpendicular on an imaginary line drawn on the liquid surface, tending 
to pull the surface apart along the line. Its unit is N rn 1 and dimensional 
formula is MT 2 . 

Experiments to demonstrate surface tension 

(i) When a painting brush is dipped into water, its hair gets 
separated from each other. When the brush is taken out of water, it 
is observed that its hair will cling together. This is because the free 
surface of water films tries to contract due to surface tension. 



Fig. 5.20 Force on 
a liquid surface 




Needle floats on water surface 



Hair clings together when brush is taken out 

Fig. 5.21 Practical examples for surface tension 



229 




Fig. 5.22 Surface 
tension based on 
molecular theory 



(ii) When a sewing needle is gently placed on water surface, it 
floats. The water surface below the needle gets depressed slightly. The 
force of surface tension acts tangentially. The vertical component of 
the force of surface tension balances the weight of the needle. 

5.5.2 Molecular theory of surface tension 

Consider two molecules P and Q as shown 
in Fig. 5.22. Taking them as centres and 
molecular range as radius, a sphere of influence 
is drawn around them. 

The molecule P is attracted in all directions 
equally by neighbouring molecules. Therefore 
net force acting on P is zero. The molecule Q is 
on the free surface of the liquid. It experiences 
a net downward force because the number of 
molecules in the lower half of the sphere is 
more and the upper half is completely outside 
the surface of the liquid. Therefore all the 
molecules lying on the surface of a liquid 
experience only a net downward force. 

If a molecule from the interior is to be brought to the surface of 
the liquid, work must be done against this downward force. This work 
done on the molecule is stored as potential energy. For equilibrium, a 
system must possess minimum potential energy. So, the free surface will 
have minimum potential energy. The free surface of a liquid tends 
to assume minimum surface area by contracting and remains in a 

state of tension like a stretched elastic 

membrane. 



5.5.3 Surface energy and surface tension 

The potential energy per unit area of 
the surface film is called surface energy. 
Consider a metal frame ABCD in which AB 
is movable. The frame is dipped in a soap 
solution. A film is formed which pulls AB 
inwards due to surface tension. If T is the 
surface tension of the film and I is the length 




Fig. 5.23 Surface energy 



230 



of the wire AB, this inward force is given by 2 x Tl . The number 2 
indicates the two free surfaces of the film. 

If AB is moved through a small distance x as shown in Fig. 5.23 
to the position A'B' , then work done is 
W = 2Tbc 

W 



Work down per unit area = 



2lx 



Surface energy 



T2lx 
2lx 





/e 



Surface energy = T 

Surface energy is numerically equal to surface tension. 

5.5.4 Angle of contact 

When the free surface of a liquid 
comes in contact with a solid, it 
becomes curved at the point of 
contact. The angle between the 
tangent to the liquid surface at the F::::::::^^^;::; 
point of contact of the liquid with the 
solid and the solid surface inside the For water For mercury 

liquid is called angle of contact. Fig. 5.24 Angle of contact 

In Fig. 5.24, QR is the tangent drawn at the point of contact Q. 
The angle PQR is called the angle of contact. When a liquid has concave 
meniscus, the angle of contact is acute. When it has a convex meniscus, 
the angle of contact is obtuse. 

The angle of contact depends on the nature of liquid and solid in 
contact. For water and glass, 9 lies between 8° and 18°. For pure water 
and clean glass, it is very small and hence it is taken as zero. The angle 
of contact of mercury with glass is 138°. 

5.5.5 Pressure difference across a liquid surface 

If the free surface of a liquid is plane, then the surface tension 
acts horizontally (Fig. 5.25a). It has no component perpendicular to 
the horizontal surface. As a result, there is no pressure difference between 
the liquid side and the vapour side. 

If the surface of the liquid is concave (Fig. 5.25b), then the resultant 



231 



excess pressure 




excess pressure 



(a) (b) (c) 

Fig. 5.25 Excess of pressure across a liquid surface 

force R due to surface tension on a molecule on the surface act vertically 
upwards. To balance this, an excess of pressure acting downward on 
the concave side is necessary. On the other hand if the surface is 
convex (Fig. 5.25c), the resultant R acts downward and there must be 
an excess of pressure on the concave side acting in the upward direction. 

Thus, there is always an excess of pressure on the concave side of 
a curved liquid surface over the pressure on its convex side due to surface 
tension. 



5.5.6 Excess pressure inside a liquid drop 

Consider a liquid drop of radius r. The molecules on the surface 
of the drop experience a resultant force acting inwards due to surface 
tension. Therefore, the pressure inside the drop must be greater than 
the pressure outside it. The excess of pressure P inside the drop provides 
a force acting outwards perpendicular to the surface, to balance the 
resultant force due to surface tension. Imagine the drop to be divided 
into two equal halves. Considering the equilibrium of the upper 

hemisphere of the drop, the upward force 
on the plane face ABCD due to excess 
pressure P is Pn r 2 (Fig. 5.26). 

If T is the surface tension of the 
liquid, the force due to surface tension 
acting downward along the circumference 
of the circle ABCD is T 2nr. 




Fig. 5.26 Excess pressure 
inside a liquid drop 



At equilibrium, 

2T 

.: P = 

r 



Pur 2 = T2nr 



232 



i.e 



P = 



Excess pressure inside a soap bubble 

A soap bubble has two liquid surfaces in contact with air, one 
inside the bubble and the other outside the bubble. Therefore the force 
due to surface tension = 2 x 2nrT 

:. At equilibrium, Pnr 2 = 2 x 2nfT 

4T 
r 
Thus the excess of pressure inside a drop is inversely proportional 

to its radius i.e. Pa - . As P a -, the pressure needed to form a very 
small bubble is high. This explains why one needs to blow hard to start 
a balloon growing. Once the balloon has grown, less air pressure is 
needed to make it expand more. 

5.5.7 Capillarity 

The property of surface tension gives rise to an interesting 
phenomenon called capillarity. When a capillary tube is dipped in water, 
the water rises up in the tube. The level of water in the tube is above 
the free surface of water in the beaker (capillary rise) . When a capillary 
tube is dipped in mercury, mercury also rises in the tube. But the level 

of mercury is depressed below the free 
surface of mercury in the beaker 
(capillary fall). 

The rise of a liquid in a capillary 
tube is known as capillarity. The height 
ft in Fig. 5.27 indicates the capillary 
rise (for water) or capillary fall (for 
mercury). 







.Vi- 


'- / 


h 





























"JB 



lT 



For water For mercury 

Fig. 5.27 Capillary rise 



Illustrations of capillarity 

(i) A blotting paper absorbs ink by capillary action. The pores in 
the blotting paper act as capillaries. 

(ii) The oil in a lamp rises up the wick through the narrow spaces 
between the threads of the wick. 

(iii) A sponge retains water due to capillary action. 

(iv) Walls get damped in rainy season due to absorption of water 
by bricks. 



233 



l«-> 




-^ R sin 9 

D 



Fig. 5.28 Surface tension 
by capillary rise method 



5.5.8 Surface tension by capillary rise method 

Let us consider a capillary tube of 
uniform bore dipped vertically in a beaker 
containing water. Due to surface tension, 
water rises to a height h in the capillary tube 
as shown in Fig. 5.28. The surface tension T 
of the water acts inwards and the reaction of Rsme <- 
the tube R outwards. R is equal to T in 
magnitude but opposite in direction. This 
reaction R can be resolved into two 
rectangular components. 

(i) Horizontal component R sin 8 acting 
radially outwards 

(ii) Vertical component R cos 8 acting 
upwards. 

The horizontal component acting all 
along the circumference of the tube cancel 
each other whereas the vertical component balances the weight of water 
column in the tube. 

Total upward force = R cos 8 x circumference of the tube 
(i.e) F = 2-kt R cos 8 or F = 2nr T cos 8 ...(1) 

[v R = T] 

This upward force is responsible for the 
capillary rise. As the water column is in 
equilibrium, this force acting upwards is 
equal to weight of the water column acting 
downwards. 

(i.e) F = W ...(2) 

Now, volume of water in the tube is 
assumed to be made up of (i) a cylindrical 
water column of height h and (ii) water in the 
meniscus above the plane CD. 

Volume of cylindrical water column = nr^h 

Volume of water in the meniscus = (Volume of cylinder of height 
r and radius r) - (Volume of hemisphere) 




D 



Fig. 5.29 Liquid 
meniscus 



234 



.*. Volume of water in the meniscus = [nr 2 x r) - I — n r 

= — nr 3. 



Total volume of water in the tube = nr^h + — nr 3 



= nr 2 h + 



If p is the density of water, then weight of water in the tube is 



• (3) 



W = nr 2 l h + 3 |P9 
Substituting (1) and (3) in (2), 



nr 2 h 



pg 



2nrT cos 6 



T = 



h + -\r P g 
2 cos e 



Since r is very small, - can be neglected compared to h. 
hrpg 



■■ 1 ~ 2cos 9 

For water, 8 is small, therefore cos 9=1 



T = 



hrpg 



5.5.9 Experimental determination 
of surface tension of water 
by capillary rise method 

A clean capillary tube of uniform 
bore is fixed vertically with its lower 
end dipping into water taken in a 
beaker. A needle N is also fixed with 
the capillary tube as shown in the Fig. 
5.30. The tube is raised or lowered until 
the tip of the needle just touches the 
water surface. A travelling microscope 
M is focussed on the meniscus of the 



JL 



(E fl~M 




Fig. 5.30 Surface tension 
by capillary rise method 



235 



water in the capillary tube. The reading R 1 corresponding to the lower 
meniscus is noted. The microscope is lowered and focused on the tip of 
the needle and the corresponding reading is taken as Rg. The difference 
between R 1 and Rg gives the capillary rise h. 

The radius of the capillary tube is determined using the travelling 
microscope. If p is the density of water then the surface tension of water 

is given by T = — - — where g is the acceleration due to gravity. 

5.5.10 Factors affecting surface tension 

Impurities present in a liquid appreciably affect surface tension. A 
highly soluble substance like salt increases the surface tension whereas 
sparingly soluble substances like soap decreases the surface tension. 

The surface tension decreases with rise in temperature. The 
temperature at which the surface tension of a liquid becomes zero is 
called critical temperature of the liquid. 

5.5.11 Applications of surf ace tension 

(i) During stormy weather, oil is poured into the sea around the 
ship. As the surface tension of oil is less than that of water, it spreads 
on water surface. Due to the decrease in surface tension, the velocity 
of the waves decreases. This reduces the wrath of the waves on the 
ship. 

(ii) Lubricating oils spread easily to all parts because of their low 
surface tension. 

(iii) Dirty clothes cannot be washed with water unless some 
detergent is added to water. When detergent is added to water, one end 
of the hairpin shaped molecules of the detergent get attracted to water 
and the other end, to molecules of the dirt. Thus the dirt is suspended 
surrounded by detergent molecules and this can be easily removed. 
This detergent action is due to the reduction of surface tension of water 
when soap or detergent is added to water. 

(iv) Cotton dresses are preferred in summer because cotton dresses 
have fine pores which act as capillaries for the sweat. 



236 



iL 



■-&-- 



3EH 



i i 



5.6 Total energy of a liquid 

A liquid in motion possesses pressure energy, kinetic energy and 
potential energy. 

(i) Pressure energy 

It is the energy possessed by a liquid 
by virtue of its pressure. 

Consider a liquid of density p contained 
in a wide tank T having a side tube near the 
bottom of the tank as shown in Fig. 5.31. A 
frictionless piston of cross sectional area 'a' 
is fitted to the side tube. Pressure exerted pig_ 5.31 Pressure energy 
by the liquid on the piston is P = h p g 

where h is the height of liquid column above the axis of the side tube. 
If x is the distance through which the piston is pushed inwards, then 

Volume of liquid pushed into the tank = ax 

.*. Mass of the liquid pushed into the tank = ax p 

As the tank is wide enough and a very small amount of liquid is 
pushed inside the tank, the height h and hence the pressure P may be 
considered as constant. 

Work done in pushing the piston through the distance x = Force 
on the piston x distance moved 

(i.e) W = Pax 

This work done is the pressure energy of the liquid of mass axp. 

r , , PCUC P 

:. Pressure energy per unit mass of the liquid = = — 

(ii) Kinetic energy 

It is the energy possessed by a liquid by virtue of its motion. 
If m is the mass of the liquid moving with a velocity u, the kinetic 
energy of the liquid = — mu 2 . 



Kinetic energy per unit mass 



1 2 

-mu 

2 



237 



(Hi) Potential energy 

It is the energy possessed by a liquid by virtue of its height above 
the ground level. 

If m is the mass of the liquid at a height h from the ground level, 
the potential energy of the liquid = mgh 

mgh 
Potential energy per unit mass = = gh 

Total energy of the liquid in motion = pressure energy + kinetic 
energy + potential energy. 

P v 2 
:. Total energy per unit mass of the flowing liquid = — + — + gh 

p z 

5.6.1 Equation of continuity 

Consider a non-viscous liquid in streamline flow through a tube 
AB of varying cross section as shown in Fig. 5.32 Let a : and o^ be the 
area of cross section, v 1 and v 2 be the 
velocity of flow of the liquid at A and B 
respectively. 

/. Volume of liquid entering per second 
at A = cijDj. 

If p is the density of the liquid, 

then mass of liquid entering per second 

at A = a,v,p. 

1 x Fig. 5.32 Equation of 

Similarly, mass of liquid leaving per continuity 

second at B = o^v^p 

If there is no loss of liquid in the tube and the flow is steady, then 
mass of liquid entering per second at A = mass of liquid leaving per 
second at B 

(i.e) cijUjp = a^p or a l v l = a z v 2 

i.e. av = constant 

This is called as the equation of continuity. From this equation 

1 

v a — . 
a 

i.e. the larger the area of cross section the smaller will be the velocity 
of flow of liquid and vice-versa. 

238 




Ground level 



5.6.2 Bernoulli's theorem ^~-^*^ 2 

In 1738, Daniel Bernoulli 
proposed a theorem for the 
streamline flow of a liquid based 
on the law of conservation of 
energy. According to Bernoulli's 
theorem, for the streamline flow of _ 
a non-viscous and incompressible 
liquid, the sum of the pressure Fig. 5.33 Bernoulli's theorem 

energy, kinetic energy and 
potential energy per unit mass is a constant. 

P v 2 

(i.e) — h + ah = constant 

p 2 y 

This equation is known as Bernoulli's equation. 

Consider streamline flow of a liquid of density p through a pipe AB 
of varying cross section. Let P 1 and P 2 be the pressures and a 1 and a 2 , 
the cross sectional areas at A and B respectively. The liquid enters A 
normally with a velocity v 1 and leaves B normally with a velocity v 2 . The 
liquid is accelerated against the force of gravity while flowing from A to 
B, because the height of B is greater than that of A from the ground 
level. Therefore P, is greater than P„. This is maintained by an external 
force. 

The mass m of the liquid crossing per second through any section 
of the tube in accordance with the equation of continuity is 

a x v x p = OjV^p = m 

m 
or cijUj = a 2 y 2 = — = V (1) 

As a 1 > a 2 , v 1 < v 2 

The force acting on the liquid at A = P l a 1 
The force acting on the liquid at B = P 2 c^ 
Work done per second on the liquid at A = P l a l x u 1 = P X V 
Work done by the liquid at B = P 2 a 2 x v i = ^2^ 
.'. Net work done per second on the liquid by the pressure energy 
in moving the liquid from A to B is = P.V - P 2 V ...(2) 



239 



If the mass of the liquid flowing in one second from A to B is m, 
then increase in potential energy per second of liquid from A to B is 
mglr^ - mg\ 

Increase in kinetic energy per second of the liquid 

1 2 * 2 

= 2 mU 2 - 2^1 

According to work-energy principle, work done per second by the 
pressure energy = Increase in potential energy per second + Increase 
in kinetic energy per second 

1 2 1 2 
2 l 



(i.e) PjV - P 2 V = (mgh^- mgh l )+\-mv 2 

PjV + mgh 1 + -mv 2 = P 2 V + mgh 2 + ^mv 2 



p i v , 1 2 P 2 V , l ., 

— — -■- " u J --" = ^— + qh„ + -" 2 

m 2 

P l . , 1 , P2 . , 1 



m 



+ fifhj +2 U i = -^ + 9h 2 + ^ v 2 



-+ ghj + g u i = y + 9 h 2 + 2 u 



P 



P 1 

or — + oh +— u 2 = constant .-.(S) 

P y 2 

This is Bernoulli's equation. Thus the total energy of unit mass of 
liquid remains constant. 

P v 2 
Dividing equation (3) by g, — + — + h = constant 

Each term in this equation has the dimension of length and hence 

P v 2 

is called head. ~ris called pressure head, — is velocity head and h is 

the gravitational head. 

Special case : 

If the liquid flows through a horizontal tube, h 1 = h^. Therefore 
there is no increase in potential energy of the liquid i.e. the gravitational 
head becomes zero. 

.*. equation (3) becomes 

— + — v 2 = a constant 
P 2 

This is another form of Bernoulli's equation. 



240 



5.6.3 Application of Bernoulli's theorem 

(i) Lift of an aircraft wing 

A section of an aircraft wing 



High velocity; Low pressure 
air flow 




Low velocity; High pressure 

Fig. 5.34 Lift of an aircraft wing 



and the flow lines are shown in 

Fig. 5.34. The orientation of the wing 

relative to the flow direction causes 

the flow lines to crowd together above 

the wing. This corresponds to 

increased velocity in this region and 

hence the pressure is reduced. But below the wing, the pressure is 

nearly equal to the atmospheric pressure. As a result of this, the upward 

force on the underside of the wing is greater than the downward force 

on the topside. Thus there is a net upward force or lift. 

(ii) Blowing of roofs 

During a storm, the roofs of huts or 
tinned roofs are blown off without any 
damage to other parts of the hut. The 
blowing wind creates a low pressure P x 
on top of the roof. The pressure P 2 under 
the roof is however greater than P r Due 
to this pressure difference, the roof is lifted 
and blown off with the wind. 




Fig. 5.35 Blowing of roofs 



<M 



Air- 



■ Air 



T 



(Hi) Bunsen burner 

In a Bunsen burner, the gas comes out of the 
nozzle with high velocity. Due to this the pressure in 
the stem of the burner decreases. So, air from the 
atmosphere rushes into the burner. 



Gas 



(iv) Motion of two parallel boats 

When two boats separated by a small distance 
row parallel to each other along the same direction, 
the velocity of water between the boats becomes very 
large compared to that on the outer sides. Because 
of this, the pressure in between the two boats gets 
reduced. The high pressure on the outer side pushes 

the boats inwards. As a result of this, the boats come closer and may 

even collide. 



Fig. 



5.36 Bunsen 
Burner 



241 



Solved problems 

5.1 A 50 kg mass is suspended from one end of a wire of length 4 m 
and diameter 3 mm whose other end is fixed. What will be the 
elongation of the wire? Take q = 7 x 10 10 N nr 2 for the material of 
the wire. 

Data :l = 4 m; d = 3 mm = 3 x 10 3 m; m = 50kg;q= 7xl0 10 N nr 2 

Fl 



Solution : q = 



Adl 
Fl 50x9.8x4 



■'■ dl m 2 q 3.14x(1.5xl0~ 3 ) 2 x7xl0 10 

= 3.96x 10- 3 m 

5.2 A sphere contracts in volume by 0.01% when taken to the bottom 
of sea 1 km deep. If the density of sea water is 10 3 kg m~ 3 , find the 
bulk modulus of the material of the sphere. 

Data :dV= 0.01% 

dV 0.01 , „ , _ q 

i.e — = y^Q- ; h = 1 km ; p = 10 J kg m J 

Solution : dP = 10 3 x 10 3 x 9.8 = 9.8 x 10 6 

. i, dP 9.8xl0 6 xl00 

■ k = dV7V= ^ = 9.8 x 10" N mi 

5.3 A hydraulic automobile lift is designed to lift cars with a maximum 
mass of 3000 kg. The area of cross-section of the piston carrying 
the load is 425 x 10 -4 m 2 . What maximum pressure would the 
piston have to bear? 

Data : m = 3000 kg, A = 425 x 1 Or 4 m 2 

Weight of car rng 
Solution: Pressure on the piston = Area oJ piston = ~J~ 

3000x9.8 
= 425x10- =6.92xlO^Nm- 2 

5.4 A square plate of 0. 1 m side moves parallel to another plate with a 
velocity of 0.1 m s -1 , both plates being immersed in water. If the 
viscous force is 2 x 10~ 3 N and viscosity of water is 10~ 3 N s m 2 , 
find their distance of separation. 



227 



Data : Area of plate A = 0.1 x 0.1 = 0.01 m 2 

Viscous force F= 2 x 10~ 3 N 

Velocity dv = 0.1 m s~ J 

Coefficient of viscosity r\ = 10~ 3 N s mr 2 

?]Adv 
Solution : Distance dx = 



F 

10 3 xO.OlxO.l 
2xl0 3 



5 x 10 4 m 



5.5 Determine the velocity for air flowing through a tube of 
10~ 2 m radius. For air p = 1 .3 kg m 3 and r| = 187 x 10~ 7 N s m 2 . 

Data:r= 10 2 m; p = 1.3 kg rn 3 ; 77= 187 x 10 7 N srn 2 ; N R = 2000 



Solution : velocity v 

17 v m 7 

= 1.44 ms 1 



P D 
2000x187 xlO 7 



1.3x2xl0 2 

5.6 Fine particles of sand are shaken up in water contained in a tall 
cylinder. If the depth of water in the cylinder is 0.3 m, calculate the 
size of the largest particle of sand that can remain suspended after 
40 minutes. Assume density of sand = 2600 kg rrr 3 and viscosity of 
water = 10~ 3 N s m 2 . 

Data :s = 0.3m, t=40 minutes = 40 x 60 s, p = 2600 kg nr 3 

Solution: Let us assume that the sand particles are spherical in 
shape and are of different size. 

Let r be the radius of the largest particle. 

0.3 . , 

Terminal velocity v = — — -— = 1.25 x 10 4 ms J 
3 40 x 60 



Radius 



9/jv 



2(p-a)g 



9xl0 3 xl.25xl0 4 
2(2600-1000)9.8 

= 5.989 x 10 6 m 



228 



5.7 A circular wire loop of 0.03 m radius is rested on the surface of a 
liquid and then raised. The pull required is 0.003 kg wt greater 
than the force acting after the film breaks. Find the surface tension 
of the liquid. 

Solution: The additional pull F of 0. 003 kg wt is the force due to 
surface tension. 

/.Force due to surface tension, 

F=Tx length of ring in contact with liquid 

(i.e) F=Tx2x2nr = 4nTr 

(i.e) 4nTr = F 

:. 4nTr = 0.003 x9.81 

0.003x9.81 
° rT= 4x3.14x0.03 = - 078JVrn 

5.8 Calculate the diameter of a capillary tube in which mercury is 
depressed by 2.219 mm. Given T for mercury is 0.54 N nr 1 , angle 
of contact is 140° and density of mercury is 13600 kg m~ 3 

Data: h = - 2.219 x 10 3 m; T = 0.54 N m 1 ; = 140° ; 

p = 13600 kg nr 3 

Solution : hrpg = 2T cos 6 





IT cos 6 




r 


hpg 






2 x 0.54 x cos 140° 




" (-2.219xlCT 3 )x 


13600x9.8 




= 2.79x1 3 m 




!r 


= 2x2.79 x 10 3 


m = 5.58 mm 



Diameter ■ 

5.9 Calculate the energy required to split a water drop of radius 
1 x 10~ 3 m into one thousand million droplets of same size. Surface 
tension of water = 0.072 N rrr 1 

Data : Radius of big drop R= 1 x 1 3 m 

Number of drops n= 10 3 x 10 6 = 10 9 ; T= 0.072 N m 1 

Solution : Let r be the radius of droplet. 



229 



Volume of 10 9 drops = Volume of big drop 

4 3 4 
W 9 X 3 nT = | nR 3 

10 9 r 3 = R 3 = (10- 3 ) 3 
(10 3 r) 3 = (10 3 ) 3 

io- 3 

Increase in surface area ds = 10 9 x 4nr 2 - 4xR 2 

(i.e) ds = 4n[ 10 9 x (1 0' 6 ) 2 - (1 3 ) 2 ] = 4n[l 3 - 1 6 / m 2 

.: ds = 0.01254 m 2 

Work doneW=T.ds = 0.072 x 0.01254 = 9.034 x 10 4 J 

5.10 Calculate the minimum pressure required to force the blood from 
the heart to the top of the head (a vertical distance of 0.5 m). Given 
density of blood = 1040 kg rrr 3 . Neglect friction. 

Data : h 2 - h x = 0.5 m , p = 1040 kg mr 3 , P 1 -P 2 = ? 

Solution : According to Bernoulli's theorem 

P 1 -P 2 = pg(h 2 -h])+-p (v 2 2 - vf) 

If v 2 = v v then 

p 1 -p 2 = pgfhz-hj 

P 1 -P 2 = 1040 x 9.8 (0.5) 
P 1 -P 2 = 5.096 x 10 3 N mr 2 



230 



Self evaluation 

(The questions and problems given in this self evaluation are only samples. 
In the same way any question and problem could be framed from the text 
matter. Students must be prepared to answer any question and problem 
from the text matter, not only from the self evaluation.) 

5. 1 If the length of the wire and mass suspended are doubled in a Young's 
modulus experiment, then, Young's modulus of the wire 

(a) remains unchanged (b) becomes double 

(c) becomes four times (d) becomes sixteen times 

5.2 For a perfect rigid body, Young's modulus is 
(a) zero (b) infinity 

(c) 1 (d) -1 

5.3 Two wires of the same radii and material have their lengths in the 
ratio 1 : 2. If these are stretched by the same force, the strains 
produced in the two wires will be in the ratio 

(a) 1 : 4 (b) 1 : 2 

(c) 2 : 1 (d) 1 : 1 

5.4 If the temperature of a liquid is raised, then its surface tension is 
(a) decreased (b) increased 

(c) does not change (d) equal to viscosity 

5.5 The excess of pressure inside two soap bubbles of diameters in the 
ratio 2 : 1 is 

(a) 1 : 4 (b) 2 : 1 

(c) 1 : 2 (d) 4 : 1 

5.6 A square frame of side I is dipped in a soap solution. When the frame 
is taken out, a soap film is formed. The force on the frame due to 
surface tension T of the soap solution is 

(a) 8TI (b)4TL 

(c) 10 TL (d) 12TI 



231 



5.7 The rain drops falling from the sky neither hit us hard nor make 
holes on the ground because they move with 

(a) constant acceleration (b) variable acceleration 

(c) variable speed (d) constant velocity 

5.8 Two hail stones whose radii are in the ratio of I : 2 fall from a height 
of 50 km. Their terminal velocities are in the ratio of 

(a) 1 : 9 (b) 9 : 1 

(c) 4 : 1 (d) 1 : 4 

5.9 Water flows through a horizontal pipe of varying cross-section at the 
rate of 0.2 m 3 s 1 . The velocity of water at a point where the area of 
cross-section of the pipe is 0.01 m 2 is 

(a) 2 ms 1 (b) 20 ms 1 

(c) 200 ms 1 (d) 0.2 ms 1 

5.10 An object entering Earth's atmosphere at a high velocity catches fire 
due to 

(a) viscosity of air (b) the high heat content of atmosphere 

(c) pressure of certain gases (d) high force ofg. 

5.11 Define : i) elastic body ii) plastic body Hi) stress iv) strain v) elastic 
limit vi) restoring force 

5.12 State Hooke 's law. 

5.13 Explain the three moduli of elasticity. 

5.14 Describe Searle 's Experiment. 

5.15 Which is more elastic, rubber or steel? Support your answer. 

5.16 State and prove Pascal's law without considering the effect of gravity. 

5.17 Taking gravity into account, explain Pascal's law. 

5.18 Explain the principle, construction and working of hydraulic brakes. 

5.19 What is Reynold's number? 

5.20 What is critical velocity of a liquid? 

5.21 Why aeroplanes and cars have streamline shape? 

5.22 Describe an experiment to determine viscosity of a liquid. 



232 



5.23 What is terminal velocity? 

5.24 Explain Stake's law. 

5.25 Derive an expression for terminal velocity of a small sphere falling 
through a viscous liquid. 

5.26 Define cohesive force and adhesive force. Give examples. 

5.27 Define i) molecular range ii) sphere of influence Hi) surface tension. 

5.28 Explain surface tension on the basis of molecular theory. 

5.29 Establish the relation between surface tension and surface energy. 

5.30 Give four examples of practical application of surface tension. 

5.31 How do insects run on the surface of water? 

5.32 Why hot water is preferred to cold water for washing clothes? 

5.33 Derive an expression for the total energy per unit mass of a flowing 
liquid. 

5.34 State and prove Bernoulli's theorem. 

5.35 Why the blood pressure in humans is greater at the feet than at the 
brain? 

5.36 Why two holes are made to empty an oil tin? 

5.37 A person standing near a speeding train has a danger of falling 
towards the train. Why? 

5.38 Why a small bubble rises slowly through a liquid whereas the bigger 
bubble rises rapidly? 

Problems 

5.39 A wire of diameter 2.5 mm is stretched by a force of 980 N. If the 
Young's modulus of the wire is 12.5 x 10 10 Nm 2 ,find the percentage 
increase in the length of the wire. 

5.40 Two wires are made of same material. The length of the first wire is 
half of the second wire and its diameter is double that of second 
wire. If equal loads are applied on both the wires, find the ratio of 
increase in their lengths. 

5.41 The diameter of a brass rod is 4 mm. Calculate the stress and strain 
when it is stretched by 0.25% of its length. Find the force exerted. 
Given q = 9.2 x 10 10 N mr 2 for brass. 



233 



5.42 Calculate the volume change of a solid copper cube, 40 mm on each 
side, when subjected to a pressure of 2 xlO 7 Pa. Bulk modulus of 
copper is 1.25 x 10 11 N mr 2 . 

5.43 In a hydraulic lift, the piston P 2 has a diameter of 50 cm and that of 
P 1 is 10 cm. What is the force on P 2 when 1 N of force is applied on 
Pl ? 

5.44 Calculate the mass of water flowing in 10 minutes through a tube of 
radius 1 2 m and length 1 m having a constant pressure of 0.2 m of 
water. Assume coefficient of viscosity of water = 9 x 10~ 4 N s mr 2 
and g = 9.8 m s~ 2 . 

5.45 A liquid flows through a pipe oflO 3 m radius and 0.1 m length 
under a pressure oflO 3 Pa. If the coefficient of viscosity of the liquid 
is 1.25 x 1 0~ 3 N s mr 2 , calculate the rate of flow and the speed of the 
liquid coming out of the pipe. 

5.46 For cylindrical pipes, Reynold's number is nearly 2000. If the diameter 
of a pipe is 2 cm and water flows through it, determine the velocity of 
the flow. Take rjfor water = 1 3 N s mr 2 . 

5.47 In a Poiseuille' s flow experiment, the following are noted, 
i) Volume of liquid discharged per minute = 15 x 10~ 6 m 3 
ii) Head of liquid = 0.30 m 

Hi) Length of tube = 0.25 m 
iv) Diameter = 2x1 0~ 3 m 

v) Density of liquid = 2300 kg mr 3 . 
Calculate the coefficient of viscosity. 

5.48 An air bubble of 0.01 m radius raises steadily at a speed of 
5x1 3 m s~ J through a liquid of density 800 kg mr 3 . Find the 
coefficient of viscosity of the liquid. Neglect the density of air. 

5.49 Calculate the viscous force on a ball of radius 1 mm moving through 
a liquid of viscosity 0.2 N s mr 2 at a speed of 0.07 m s" 1 . 

5.50 A U shaped wire is dipped in soap solution. The thin soap film formed 
between the wire and a slider supports a weight of 1.5 x 10 2 N. If 
the length of the slider is 30 cm, calculate the surface tension of the 
film. 



234 



5.51 Calculate the force required to remove aflat circular plate of radius 
0.02 mfrom the surface of water. Assume surface tension of water 
is 0.07 Nm 1 . 

5.52 Find the work done in blowing up a soap bubble from an initial surface 
area of 0.5 x 1 4 m 2 to an area l.lxl 4 m 2 . The surface tension of 
soap solution is 0.03 N m 1 . 

5.53 Determine the height to which water will rise in a capillary 
tube of 0.5 x 10 3 m diameter. Given for water, surface tension 
is 0.074 Nm 1 . 

5.54 A capillary tube of inner diameter 4 mm stands vertically in a bowl 
of mercury. The density of mercury is 13,500 kg m 3 and its surface 
tension is 0.544 N m 1 . If the level of mercury in the tube is 2.33 mm 
below the level outside, find the angle of contact of mercury with 
glass. 

5.55 A capillary tube of inner radius 5 x 10 4 m is dipped in water of 
surface tension 0.075 N m 1 . To what height is the water raised by 
the capillary action above the water level outside. Calculate the weight 
of water column in the tube. 

5.56 What amount of energy will be liberated if 1000 droplets of water, 
each of diameter 1 8 m, coalesce to form a big drop. Surface tension 
of water is 0.075 N mr 1 . 

5.57 Water flows through a horizontal pipe of varying cross-section. If the 
pressure of water equals 2 x 10 2 m of mercury where the velocity of 
flow is 32 x 1 2 m s^ 1 find the pressure at another point, where the 
velocity of flow is 40 x 1 0~ 2 m s^ 1 . 



235 







Answers 




5.1 (a) 


5.2 (b) 


5.3 (d) 


5.4 (a) 


5.5 (c) 


5.6 (a) 


5.7 (d) 


5.8 (d) 


5.9 (b) 


5.10 (a) 






5.39 


0.16% 


5.40 


1 :8 


5.41 


2.3 x 10 8 Nm 2 , 0.0025. 2.89 x 10 3 N 


5.42 


-1.024 x 10' 8 m 3 


5.43 


25 N 


5.44 


5.13 x 10 3 kg 


5.45 


3.14 x lO^mPs' 1 , 1 ms 1 


5.46 


0.1 ms 1 


5.47 


4.25 x 10' 2 Nsm 2 


5.48 


34.84 N s m 2 


5.49 


2.63 x 10 4 N 


5.50 


2.5 x lO^Nm 1 


5.51 


8.8 x 10 3 N 


5.52 


1.8 x 10 6 J 


5.53 


6.04 x 10 2 m 


5.54 


124°36' 


5.55 


3.04 x 10 2 m. 2.35 x 10 4 N 


5.56 


2.12 x 10 14 J 


5.57 


2636.8 Nm 2 



236 



Mathematical Notes 

(Not for examination) 
Logarithm 

In physics, a student is expected to do the calculation by using 
logarithm tables. The logarithm of any number to a given base is the 
power to which the base must be raised in order to obtain the number. 
For example, we know that 2 raised to power 3 is equal to 8 (i.e) 2 3 = 8. In 
the logarithm form this fact is stated as the logarithm of 8 to the base 2 is 
equal to 3. (i.e.) log 2 8 = 3. 

In general, if a* = N, then log a N = x. 

We use "common logarithm" for calculation purposes. Common 
logarithm of a number is the power to which 10 must be raised in order to 
obtain that number. The base 10 is usually not mentioned. In other words, 
when base is not mentioned, it is understood as base of 10. 

For doing calculations with log tables, the following formulae should 
be kept in mind. 

(i) Product formula : log mn = log m + log n 

m 
(ii) Quotient formula : log — = log m - log n 

(iii) Power formula : log m n = n log m 

(iv) Base changing formula : log a m = log b m x log a b 

Logarithm of a number consists of two parts called characteristic 
and Mantissa. The integral part of the logarithm of a number after 
expressing the decimal part as a positive is called characteristic. The 
positive decimal part is called Mantissa. 

To find the characteristic of a number 

(i) The characteristic of a number greater than one or equal to 
one is lesser by one (i.e) (n - 1) than the number of digits (n) 
present to the left of the decimal point in a given number. 

(ii) The characteristic of a number less than one is a negative 
number whose numerical value is more by one i.e. (n+1) than 



the number of zeroes (n) between the decimal point and the 
first significant figure of the number. 
Example Number Characteristic 

5678.9 3 

567.89 2 

56.789 1 

5.6789 _0_ 

0.56789 J_ 

0.056789 _2_ 

0.0056789 3 

To find the Mantissa of a number 

We have to find out the Mantissa from the logarithm table. The 
position of a decimal point is immaterial for finding the Mantissa, 
(i.e) log 39, log 0.39, log 0.039 all have same Mantissa. We use the following 
procedure for finding the Mantissa. 

(i) For finding the Mantissa of log 56.78, the decimal point is ignored. 
We get 5678. It can be noted that the first two digits from the left 
form 56, the third digit is 7 and the fourth is 8. 

(ii) In the log tables, proceed in the row 56 and in this row find the 
number written under the column headed by the third digit 7. 
(i.e) 7536. To this number the mean difference written under 
the fourth digit 8 in the same row is added (i.e) 
7536 + 6 = 7542. Hence logarithm of 56.78 is 1.7542. 1 is the 
characteristic and 0.7542 is the Mantissa. 

(iii) To find out the Mantissa of 567, find the number in the row 
headed by 56 and under the column 7. It is 7536. Hence the 
logarithm of 567 is 2. 7536. Here 2 is the characteristic and 
0.7536 is the Mantissa. 

(iv) To find out the Mantissa of 56, find the number in the row 
headed by 56 and under the column 0. It is 7482. Hence the 
logarithm of 56 is 1.7482. Here 1 is the characteristic and 0.7482 
is the Mantissa. 

(v) To find out the Mantissa of 5, find the number in the row headed 
by 50 and under the column 0. It is 6990. Hence the logarithm 



of 5 is 0.6990. Here is the characteristic and 0.6990 is the 
Mantissa. 

Antilogarithm 

To find out the antilogarithm of a number, we use the decimal part 
of a number and read the antilogarithm table in the same manner 
as in the case of logarithm, 
(i) If the characteristic is n, then the decimal point is fixed after 

(n+l)th digit, 
(ii) If the characteristic is n, then add (n-1) zeroes to the left side 

and then fix the decimal point, 
(iii) In general if the characteristic is n or n, then fix the decimal 

point right side of the first digit and multiply the whole number 

by 10 n or 10 n . 



Example 



Number 

0.9328 
1.9328 
2.9328 
3.9328 
T.9328 
"2.9328 
3.9328 



Antilogarithm 

8.567 or 8.567 x 10° 
85.67 or 8.567 x 10 1 
856.7 or 8.567 x 10 2 
8567.0 or 8.567 x 10 3 
0.8567 or 8.567 x 10" 1 
0.08567 or 8.567 x 10 2 
0.008567 or 8.567 x 10 3 



EXERCISE - 1 

Expand by using logarithm formula 



(i) 



(iii) 



T= 2ti 



nr 2 x 



2. Multiply 5 .5670 by 3 



(ii) v e = ^2gR 
(iv) log e 2 



3. Divide 3 .6990 by 2 

4. Evaluate using logarithm 
2x22x6400 



(i) 



7x7918.4 



(ii) V9.8x6370xl0 3 



2x7.35xl0 2 0.5 

9.8xltfx8.5xia 2 (1V) 2n V245 



Some commonly used formulae of algebra 

(i) (a+b) 2 = a 2 + 2ab + b 2 

(ii) (a-b) 2 = a 2 - 2ab + b 2 

(iii) (a+b+c) 2 = a 2 + b 2 + c 2 + 2ab + 2bc + 2ca 

(iv) (a+b) 3 = a 3 + b 3 + 3a 2 b + 3b 2 a 

(v) (a-b) 3 = a 3 - b 3 - 3a 2 b + 3ab 2 

Quadratic equation 

An algebraic equation in the form ax? + bx + c = is called quadratic 
equation. Here a is the coefficient of x 2 , b is the coefficient of x and c is the 
constant. The solution of the quadratic equation is 



x = 



-b ± y/b 2 - 4ac 



la 

Binomial theorem 

n[n - 1) „ 
The theorem states that (1 + x) n = 1 + nx + — — — x z + 

n(n-l)(n-2) , 

— x 3 + ... where x is less than 1 and n is any number. If n 

is a positive integer the expansion will have (n+1) terms and if n is 
negative or fraction, the expansion will have infinite terms. 

Factorial 2 = 2! = 2 x 1 
Factorial 3 = 3! = 3x2x1 
Factorial n = n\ = n(n - 1) (n - 2) .... 

If x is very small, then the terms with higher powers of x can be 
neglected. 

(i.e) (1 + x) n = 1 + nx (1 + x) n = 1 - nx 

(1 - x) n = 1 - nx (1 - x) n = 1 + nx 

EXERCISE- 2 

1 . Find the value of x in Ax 2 + 5x - 2 = 

1-2 



2. Expand Binomially (i) 



1- 



(ii) (1 - 2x) 3 



Trigonometry 

Let the line AC moves in anticlockwise direction from the initial 
position AB. The amount of revolution that the moving line makes with its 
initial position is called angle. From the figure 8 = [CAB . The angle is 
measured with degree and radian. Radian is the angle subtended at the 
centre of a circle by an arc of the circle, whose length is equal to the radius 
of the circle. 

1 radian = 57° 17' 45" 1 right angle = tt/2 radian 

1° = 60' (sixty minutes). 1' = 60" (sixty seconds) 



Triangle laws of sine and cosine 





z 2 - 2bc cos a 



a 
sin a 



Trigonometrical ratios (T - ratios) 

Consider the line OA making an 
angle 8 in anticlockwise direction with 
OX. 

From A, draw the perpendicular 
AB to OX. 

The longest side of the right angled 
triangle, OA is called hypotenuse. The side 
AB is called perpendicular or opposite side. 
The side OB is called base or adjacent side. 



b 
sin/? 



c 

sin y 



X' 



O 



Y' 




1 . Sine of angle 8 = sin 8 = 



2. Cosine of angle 8 = cos 8 



perpendicular 
hypotenuse 

base 
hypotenuse 



B 



X 



3. Tangent of angle 8 = tan 8 = 



perpendicular 



base 



4. Cotangent of 8 = cot 8 = 



base 



perpendicular 



5. Secant of 8 = sec 8 = 



hypotenuse 



6. Cosecant of 8 = cosec 8 = 



base 

hypotenuse 



perpendicular 



Sign of trigonometrical ratios 



II quadrant 

sin 8 and cosec 8 
only positive 


I quadrant 

All positive 


III quadrant 

tan 8 and cot 8 
only positive 


IV quadrant 

cos 8 and sec 8 only 
positive 



T - ratios of allied angles 

- 8, 90° - 6, 90° + 8, 180° - 8, 180° + 8, 270° - 8, 270° + 8 are called 
allied angles to the angle 8. The allied angles are always integral multiples 
of 90°. 



1 . (a) sir (-9) = - sir I 



(b) cos (-9) = cos i 



2. (a) sir (90° - 6) = cos 9 (b) cos (98° - 6) = sir i 



3. (a) sir (90° + 6) = cos 9 (b) cos (90° + 9) = - sir 9 (c 



4. (a) sir (180° - 9) = sir 9 (b) cos (180° - 9) = - cos 9 (c 



5. (a)siR(180° + 9) = -siR9 (b) cos (180° + 9) =- cos 9 (c 



6. (a) sir (270° - 9) = - cos 9 (b) cos (270° 



SIR 9 IC 



7. (a) sir (270° + 9) = - cos 9 (b) cos (270° + 9) = sir 9 (c 



tan (-9) = - taR 



tan (90° - 9) = cot i 



taR (98° + 9)=- cot ( 



tan (180°-9) = - tan l 



taR (180° + 9) = taR i 



taR (270° - 9) = cot i 



taR (270° + 9)= - cot 



6 



T- ratios of some standard angles 



Angle 


0° 


30° 


45° 


60° 


90° 


120° 


180° 


sin 8 





1 
2 


1 

12 


2 


1 


73" 
2 





cos 8 


1 


2 


1 


1 
2 





1 

~ 2 


-1 


tan 8 





1 

73 


1 


V3 


oo 


-V3 






Some trigonometric formulae 



1. 

2. 
3. 
4. 

5. 
6. 

7. 
8. 

9. 

10. 
11. 

12. 
13. 
14. 



sin (A + B) = sin A cos B + cos A sin B 

cos (A + B) = cos A cos B - sin A sin B 

sin (A - B) = sin A cos B - cos A sin B 

cos (A - B) = cos A cos B + sin A sin B 

A+B A-B 

sin A + sin B = 2 sin — ; — cos — ; — 

A+B A-B 

sin A - sin B = 2 cos — ; — sin 

A+B 
cos A + cos B = 2 cos — ; — COS 

A+B A- 

cos A - cos B = 2 sin — ; — sin — y 

2 tan A 



2 
A-l 



sin 2A = 2 sin A cos A = : r~iT 

1 + tan 2 A 

2 sin A cos B = sin (A + B) + sin (A - B) 
2 cos A sinB = sin (A + B) - sin (A - B) 
2 sin AsinB = cos (A-B) - cos (A + B) 
2 cos A cos B = cos (A + B) + cos (A - B) 
cos 2A= 1 - 2 sin 2 A 



Differential calculus 

Let y be the function of x 
(i.e) y = f(x) 



• (1) 



The function y depends on variable x. If the variable x is changed to 
x + Ax, then the function is also changed to y + Ay 



.-. y + Ay = f [x + Ax) ....(2) 

Subtracting equation (1) from (2) 
Ay = f (x + Ax ) - f (x ) 
dividing on both sides by Ax, we get 

Ay __ f(x+Ax)-f(x) 

AX AX 

Taking limits on both sides of equation, when Ax approaches zero, 
we get 



Lt 

Ax^>0 



^Ay^_ f (x+Ax)- f (x) 



Ax J ax^o Ax 



. A} dy 

In calculus L t — is denoted by — and is called differentiation of 

ax^o ax J dx 

y with respect to x. 

The differentiation of a function with respect to a variable means 
the instantaneous rate of change of the function with respect to the 
variable. 

Some theorems and formulae 

d 

1 . -r- (c) = , if c is a constant. 

dx 

2. If y = c u, where c is a constant and u is a function of x then 

dy d du 

— = — (cu) = c — 
dx dx dx 

3. Ify=U +V+W where u, v and w are functions of x then 

dy d , , du dv dw 

— = — (u±v±w)= ± ± 

dx dx dx dx dx 

4. If y = x 11 , where n is the real number then 

^=^(x") = nx- 
dx dx 

5. If y = uu where u and v are functions of x then 

dy d , . dv du 
-/- = — (uv)=u— + v — 
dx dx dx dx 



8 



d 
dx 


(e x ) 


= e x 




d 
dx 


(log. 


x) = 


1 

X 



dy 
6. If u is a function of x, then du = — . dx 

y y dx 

d . .. . 
7. 

8. 

d 

9. "ttt. {sin 8) = cos 8 
iff 

d 

10. — - (cos 8) = - sin 8 
iff 

11. If y is a trigonometrical function of 8 and 8 is the function of t, then 

d iff 

tt {sin 8) = cos 8 -rr 
dt dt 

12. If y is a trigonometrical function of 8 and 8 is the function of t, then 

_d_ iff 

jjj. (cos 8) = - sin8 -77- 

EXERCISE - 3 

dy 

1. If y = sin 38 find -77- 

d8 

dy 

2. If y = x 5 / 7 find -7^ 

y dx 

1 dy 

3. If y = -7 find — 

x dx 

dy 

4. If y = 4X 3 + 3X 2 + 2, find -^ L 

dx 

5. Differentiate : (i) ax 2 + bx + c 

f ds l 

6. If s = 2tP - 5t 2 + 4t - 2, find the position (s), velocity ~rr and 

fdv^ 
acceleration — of the particle at the end of 2 seconds. 

VdtJ 

Integration 

It is the reverse process of differentiation. In other words integration 
is the process of finding a function whose derivative is given. The integral 
of a function y with respect to x is given by y dx. Integration is represented 
by the elongated S. The letter S represents the summation of all differential 
parts. 



Indefinite integral 

We know that 



dx 



dx 
_d_ 

dx 



3x 2 
3 +4) =3x 2 



x 3 +c) 



3x 



The result in the above three equations is the same. Hence the 
question arises as to which of the above results is the integral of 3X 2 . To 
overcome this difficulty the integral of 3x? is taken as (x 3 + c) , where c is 
an arbitrary constant and can have any value. It is called the constant of 
integration and is indefinite. The integral containing c, (i.e) (x 3 + c) is 
called indefinite integral. In practice 'c' is generally not written, though it 
is always implied. 

Some important formulae 

(1) \dx=x 



(2) Jx'd) 



n+l 



dx 

_d_ 
dx 



r t ^\ 



n+l 



X 11 



(3) Jcu dx = C Ju dx where c is a constant 

(4) J(u ±v ±w) dx = j"u dx + [v dx ±[w dx 

(5) \—dx = log* 
J x 

(6) jVdx = e x 

(7) fcos^d^ = sin 6> 

(8) j"sin6>d6> = -cos<9 

Definite integrals 

When a function is integrated between a lower limit and an upper 
limit, it is called a definite integral. 



10 



Jf'(x)dx=[f (x)] =f(b)-f(a) is a definite integral. Here a and b 
a a 

lower and upper limits of the variable x. 

EXERCISE - 4 

1 . Integrate the following with respect to x 

(i) 4X 3 (ii) — (iii) Sx 2 + 7x - 4 

5 2 

(iv) r-277- (v) -~f (vi) 12a 2 + 6x 

IK A 

2. Evaluate 

3 4 



are 



i, Jx 2 dx (ii)jV^dx 

2 1 

4 ff/2 

m) Jxdx (lv) | cos^d^ 



2 -»■/ 2 

ANSWERS 



Exercise - 1 



1. (i) tog 2 + tog 3. 14 + - log I - - log g 

1 
(ii) - (log 2 + logg + log R) 

(iii) log m + log g + log I - log 3.14-2 log r - log x 
(iv) 0.6931 

2. 14.7010 

3. 2.8495 

4. (i) 5.080 (ii) 7.9 x 10 3 
(iii) 1.764 x 10" 4 (iv) 2.836 x 10 1 

11 



Exercise - 2 



-5 ± V57" , 2 h 

(1) jp— (2) & l ~Y (ii)l-6x 



Exercise - 3 



5 -on "2 

(1)3 cos 36 (2) yX '" (3)— (4)12x 2 + 6x 

(5) 2ax + b (6) 2, 8, 14 



Exercise - 4 



1 3 7 2 , 

1. (i)x 4 (ii) — (in) x 3 +-x' -4x 

A Z 



(iv) x 5 / 7 (v) — (vi) 4X 3 + 3X 2 

A 

19 14 

2. (1) y (11) y (ill) 6 (iv) 2 



12 



ANNEXURE 

(NOT FOR EXAMINATION) 

Proof for Lami's theorem. 

Let forces P, Q and R acting at a point O be in equilibrium. Let 
OA and OB(=AD) represent the forces P and in magnitude and 
direction. By the parallelogram law of forces OD will represent the 
resultant of the forcesP and Q. Since the forces are in equilibrium DO 
will represent the third force R. 

In the triangle OAD, using law of sines, 

OA AD OD 



i.el 





D 


sin \ODA sin 

From Fig. 2.35, 
ZODA = ZBOD 


\AOD 
= 180° 


sin \OAD 


\ 


\ 
\ 
\ 
\ 

— — ^ 

P A\ 
\ 

\ 


- ZBOC 






ZAOD = 180° - 


ZAOC 




Proof for Lami's 


theorem 


ZOAD = 180° - 


ZAOB 





Therefore, 

OA AD OD 



sin (180° - ZBOC) sin (180° - ZAOC) sin (180° - ZAOB) 

OA AD OD 

sin ZBOC ~ sin ZAOC ~ sin ZAOB 

If ZBOC =a, ZAOC=j5, ZAOB=y 

P Q R 

sin a sin fi sin y whlch P roves Lami ' s theorem - 



237 



1 . Moment of inertia of a thin uniform rod 



(i) About an axis passing through its 
centre of gravity and perpendicular 
to its length 

Consider a thin uniform rod AB of 

mass M and length I as shown in 

Fig. 1. Its mass per unit length will 

M 
be — . Let, YY' be the axis passing 

through the centre of gravity G of the 
rod (and perpendicular to the 
length AB). 

Consider a small element of length 
dx of the rod at a distance x from G. 

The mass of the element 



jy, ; y 

dx 

JA ]G — *lk— 


■ 




1 1 x 


i i i 

1 2 1 



]B 



Fig 1 Moment of inertia of 
a thin uniform rod 



M 



= mass per unit length x length of the element = -r- x dx ...(1) 
The moment of inertia of the element dx about the axis YY' is, 
dl = (mass) x (distance) 2 =f— dxj(x 2 ) ...(2) 

Therefore the moment of inertia of the whole rod about YY' is obtained 

I I 

by integrating equation (2) within the limits - — to + — . 



XG" 



■dx 



J x 2 dx 



-1/2 



X CG 



f-T" 2 

V ^ )-\n 

|3 13 



M 
31 



^CG _ 



31 
Ml 3 



Ml 2 
T2~ 



.(3) 



238 



(ii) About an axis passing through the end and perpendicular to its 
length 

The moment of inertia J about a parallel axis Y..Y.,' passing through 
one end A can be obtained by using parallel axes theorem 

\2 „ „2 n/n 2 



Ml 2 



MT Mf 
12 4 



2 Moment of inertia of a thin circular ring 

(i) About an axis passing through its centre and 
perpendicular to its plane 

Let us consider a thin ring of mass M and 
radius R with O as centre, as shown in Fig. 2. As the 
ring is thin, each particle of the ring is at a distance 
R from the axis XOY passing through O and 
perpendicular to the plane of the ring. 

For a particle of mass m on the ring, its moment 
of inertia about the axis XOY is mR 2 . Therefore the moment of inertia of 
the ring about the axis is, 

I = E mR 2 = (Em ) R 2 = MR 2 




Fig 2 Moment of 
Inertia of a ring 



(ii) About its diameter 

AB and CD are the diameters of the ring 
perpendicular to each other (Fig. 3). Since, the ring 
is symmetrical about any diameter, its moment of 
inertia about AB will be equal to that about CD. Let 
it be I. . If lis the moment of inertia of the ring about 
an axis passing through the centre and 
perpendicular to its plane then applying 
perpendicular axes theorem, 

.-. I = I d + I d = MR 2 (or) I d = -MR 2 




Fig 3 Moment of 

inertia of a ring 

about its diameter 



239 



(Hi) About a tangent 



The moment of inertia of the ring about a tangent EF parallel to AB 
is obtained by using the parallel axes theorem. The moment of inertia of 
the ring about any tangent is, 



It 



I.+MR 2 = -MR 2 
2 



-MR 2 




Fig 4 Moment of inertia of a 
circular disc 



I T = -MR 2 
T 2 

3 Moment of inertia of a circular disc 

(i) About an axis passing through its 
centre and perpendicular to its plane 

Consider a circular disc of mass M 
and radius R with its centre at O as 
shown in Fig. 4. Let a be the mass per 
unit area of the disc. The disc can be 
imagined to be made up of a large number 
of concentric circular rings of radii 
varying from O to R .Let us consider one 
such ring of radius r and width dr. 

The circumference of the ring = 2nr. 
The area of the elementary ring = 2nr dr 
Mass of the ring= 2nr dr a = 2nra dr ...(1) 

Moment of inertia of this elementary ring about the axis passing 
through its centre and perpendicular to its plane is 

dl = mass x ( distance ) 2 

= {2nra dr) r 2 ...(2) 

The moment of inertia of the whole disc about an axis passing 
through its centre and perpendicular to its plane is, 

R R 

I = j 2nav 3 dr = 2na [ r 3 dr = 2na 

o o 

27io-R i i „, \ 1 „, 1 



or 



I = 



= K-)^ 



4 v >2 

where M = nR 2 a is the mass of the disc. 



R z = 



MR 1 



.(3) 



240 



(ii) About a diameter 

Since, the disc is symmetrical about 
any diameter, the moment of inertia about 
the diameter AB will be same as its moment 
of inertia about the diameter CD. Let it be 
J d (Fig. 5). According to perpendicular axes 
theorem, the moment of inertia I of the disc, 
about an axis perpendicular to its plane and 
passing through the centre will be equal to 
the sum of its moment of inertia about two 
mutually perpendicular diameters AB 
and CD. 



Hence, I = I d + I d = 



1 



MR 2 = \ MR 2 




B 



E D F 

Fig 5 Moment of inertia of a 

disc about a tangent line 



(Hi) About a tangent in its plane 

The moment of inertia of the disc about the tangent EF in the plane 
of the disc and parallel to AB can be obtained by using the theorem of 
parallel axes (Fig. 3.15). 

I. + MR 2 = - 

a 4 

5 



Ir 



-MR 2 +MR 2 



MR 2 



4 Moment of inertia of a sphere 

(i) About a diameter 

Let us consider a homogeneous solid 
sphere of mass M, density p and radius R 
with centre O (Fig. 6). AB is the diameter 
about which the moment of inertia is to be 
determined. The sphere may be considered A 
as made up of a large number of coaxial 
circular discs with their centres lying on 
AB and their planes perpendicular to AB. 
Consider a disc of radius PO' = y and 
thickness dx with centre O' and at a 
distance x from O, 




Fig 6 Moment of inertia of a 
sphere about a diameter 



241 



...(1) 

.-(2) 
...(3) 



Its volume = ny 2 dx 

Mass of the disc = n y 2 dx . p 

From Fig. 6, R 2 = y 2 + x 2 (or) y 2 = R 2 - x 2 

Using (3) in (2), 

Mass of the circular disc = n ( R 2 - x 2 ) dx p . . . (4) 

The moment of inertia of the disc about the diameter AB is, 



dl = — (mass) x (radius) 2 



-x{R 2 -x 2 )dx.p(yf 



i^p(R 2 -x 2 ) 2 dx 



... (5) 



The moment of inertia of the entire sphere about the diameter AB is 
obtained by integrating eqn (5) within the limits x = -R to x = + R. 



I= \-MR 2 -x 2 fdx 

-R 2 

I = 2x Uxp)j(R 2 - x 2 fdx 



(xp)\(R 



2R 2 x 2 )dx 



= x P 



R" 



R 5 2R 5 



= n p I — R: 
' 15 



- M.\-R 2 
.5 



7lR? P 



-R' 




-MR 2 



Fig 7 Moment of inertia of a 
sphere about a tangent 



4 3 
where M = — nR P = mass of the solid sphere 
o 

• j= -MR 2 
5 

(ii) About a tangent 

The moment of inertia of a solid sphere about a tangent EF parallel 
to the diameter AB (Fig. 7) can be determined using the parallel axes 
theorem, 

242 



I T = Ij^ + MR 2 = ^MR 2 + MR 2 

.-. /„=- MR 2 

1 5 

5. Moment of inertia of a solid cylinder 

(i) about its own axis 

Let us consider a solid cylinder of mass M, radius R and length I. It 
may be assumed that it is made up of a large number of thin circular 
discs each of mass m and radius R placed one above the other. 

Moment of inertia of a disc about an axis passing through its centre 
but perpendicular to its plane = 



mR 2 



2 „ 2 

mR 2 



Moment of inertia of the cylinder about its axis 1=1,- 

r* f \ r 2 MR 2 



2 {^ ) 2 2 

(ii) About an axis passing through its centre and perpendicular to 
its length 

M 
Mass per unit length of the cylinder = — ■■■{I) 

Let O be the centre 

Y V. 
of gravity of the cylinder 

and YOY' be the axis 

passing through the 

centre of gravity and x / M 

perpendicular to the W 

length of the cylinder 

(Fig. 8). 

Consider a small Y ' Yi 

circular disc of width dx Fi 9-8 Moment of inertia of a 
at a distance x from the c ^ linder about its curis 

axis YY'. 

.•. Mass of the disc = mass per unit length x width 

-J dx ...(2) 

243 



Moment of inertia of the disc about an axis parallel to YY' (i.e) about 

radius 2 
its diameter = (mass) 



T dx ) (^ ) = ^f dx 



.(3) 



By parallel axes theorem, the moment of inertia of this disc about 
an axis parallel to its diameter and passing through the centre of the 
cylinder (i.e. about YY') is 



[MR 2 



M 



df = [- ir Jdx+ [yd* J (x ?) 
Hence the moment of inertia of the cylinder about YY' is, 



.(4) 



I 



C I MR 2 , M 
_ I | dx + — x dx 



41 



I 



MR 



2 +1/2 ., +1/2 

1V1JX r , Mr 2 , 

r_ dx-\ X dx 

1 - 41 J I J 

11 -1/2 L -1/2 



MR 2 r 1+ 

T- \ x \ 

1- 4l L J 



ai M 

1/2 + T 



r x 3\ +l/2 



V 3 V-i/2 



MR 2 



1= 41 



M 



41 V ; 



2l 3 
24 



MR 2 Ml 2 

+ ■ 



12 



I = M 



^1 i!_ 

4 12 



.(5) 



244